- 2.4 Solve Mixture and Uniform Motion Applications
- Introduction
- 1.1 Use the Language of Algebra
- 1.2 Integers
- 1.3 Fractions
- 1.4 Decimals
- 1.5 Properties of Real Numbers
- Key Concepts
- Review Exercises
- Practice Test
- 2.1 Use a General Strategy to Solve Linear Equations
- 2.2 Use a Problem Solving Strategy
- 2.3 Solve a Formula for a Specific Variable
- 2.5 Solve Linear Inequalities
- 2.6 Solve Compound Inequalities
- 2.7 Solve Absolute Value Inequalities
- 3.1 Graph Linear Equations in Two Variables
- 3.2 Slope of a Line
- 3.3 Find the Equation of a Line
- 3.4 Graph Linear Inequalities in Two Variables
- 3.5 Relations and Functions
- 3.6 Graphs of Functions
- 4.1 Solve Systems of Linear Equations with Two Variables
- 4.2 Solve Applications with Systems of Equations
- 4.3 Solve Mixture Applications with Systems of Equations
- 4.4 Solve Systems of Equations with Three Variables
- 4.5 Solve Systems of Equations Using Matrices
- 4.6 Solve Systems of Equations Using Determinants
- 4.7 Graphing Systems of Linear Inequalities
- 5.1 Add and Subtract Polynomials
- 5.2 Properties of Exponents and Scientific Notation
- 5.3 Multiply Polynomials
- 5.4 Dividing Polynomials
- Introduction to Factoring
- 6.1 Greatest Common Factor and Factor by Grouping
- 6.2 Factor Trinomials
- 6.3 Factor Special Products
- 6.4 General Strategy for Factoring Polynomials
- 6.5 Polynomial Equations
- 7.1 Multiply and Divide Rational Expressions
- 7.2 Add and Subtract Rational Expressions
- 7.3 Simplify Complex Rational Expressions
- 7.4 Solve Rational Equations
- 7.5 Solve Applications with Rational Equations
- 7.6 Solve Rational Inequalities
- 8.1 Simplify Expressions with Roots
- 8.2 Simplify Radical Expressions
- 8.3 Simplify Rational Exponents
- 8.4 Add, Subtract, and Multiply Radical Expressions
- 8.5 Divide Radical Expressions
- 8.6 Solve Radical Equations
- 8.7 Use Radicals in Functions
- 8.8 Use the Complex Number System
- 9.1 Solve Quadratic Equations Using the Square Root Property
- 9.2 Solve Quadratic Equations by Completing the Square
- 9.3 Solve Quadratic Equations Using the Quadratic Formula
- 9.4 Solve Equations in Quadratic Form
- 9.5 Solve Applications of Quadratic Equations
- 9.6 Graph Quadratic Functions Using Properties
- 9.7 Graph Quadratic Functions Using Transformations
- 9.8 Solve Quadratic Inequalities
- 10.1 Finding Composite and Inverse Functions
- 10.2 Evaluate and Graph Exponential Functions
- 10.3 Evaluate and Graph Logarithmic Functions
- 10.4 Use the Properties of Logarithms
- 10.5 Solve Exponential and Logarithmic Equations
- 11.1 Distance and Midpoint Formulas; Circles
- 11.2 Parabolas
- 11.3 Ellipses
- 11.4 Hyperbolas
- 11.5 Solve Systems of Nonlinear Equations
- 12.1 Sequences
- 12.2 Arithmetic Sequences
- 12.3 Geometric Sequences and Series
- 12.4 Binomial Theorem

## Learning Objectives

By the end of this section, you will be able to:

- Solve coin word problems
- Solve ticket and stamp word problems
- Solve mixture word problems
- Solve uniform motion applications

## Be Prepared 2.10

Before you get started, take this readiness quiz.

Simplify: 0.25 x + 0.10 ( x + 4 ) . 0.25 x + 0.10 ( x + 4 ) . If you missed this problem, review Example 1.52 .

## Be Prepared 2.11

The number of adult tickets is three more than twice the number of children tickets. Let c represent the number of children tickets. Write an expression for the number of adult tickets. If you missed this problem, review Example 1.11 .

## Be Prepared 2.12

Convert 4.2% to a decimal. If you missed this problem, review Example 1.40 .

Solve Coin Word Problems

Using algebra to find the number of nickels and pennies in a piggy bank may seem silly. You may wonder why we just don’t open the bank and count them. But this type of problem introduces us to some techniques that will be useful as we move forward in our study of mathematics.

If we have a pile of dimes, how would we determine its value? If we count the number of dimes, we’ll know how many we have—the number of dimes. But this does not tell us the value of all the dimes. Say we counted 23 dimes, how much are they worth? Each dime is worth $0.10—that is the value of one dime. To find the total value of the pile of 23 dimes, multiply 23 by $0.10 to get $2.30.

The number of dimes times the value of each dime equals the total value of the dimes.

This method leads to the following model.

## Total Value of Coins

For the same type of coin, the total value of a number of coins is found by using the model

- number is the number of coins
- value is the value of each coin
- total value is the total value of all the coins

If we had several types of coins, we could continue this process for each type of coin, and then we would know the total value of each type of coin. To get the total value of all the coins, add the total value of each type of coin.

## Example 2.39

Jesse has $3.02 worth of pennies and nickels in his piggy bank. The number of nickels is three more than eight times the number of pennies. How many nickels and how many pennies does Jesse have?

## Try It 2.77

Jesse has $6.55 worth of quarters and nickels in his pocket. The number of nickels is five more than two times the number of quarters. How many nickels and how many quarters does Jesse have?

## Try It 2.78

Elane has $7.00 total in dimes and nickels in her coin jar. The number of dimes that Elane has is seven less than three times the number of nickels. How many of each coin does Elane have?

The steps for solving a coin word problem are summarized below.

## Solve coin word problems.

- Determine the types of coins involved.
- Label the columns “type,” “number,” “value,” and “total value.”
- List the types of coins.
- Write in the value of each type of coin.

Step 2. Identify what you are looking for.

- Use variable expressions to represent the number of each type of coin and write them in the table.
- Multiply the number times the value to get the total value of each type of coin.
- It may be helpful to restate the problem in one sentence with all the important information. Then, translate the sentence into an equation.
- Write the equation by adding the total values of all the types of coins.
- Step 5. Solve the equation using good algebra techniques.

Step 6. Check the answer in the problem and make sure it makes sense.

Step 7. Answer the question with a complete sentence.

Solve Ticket and Stamp Word Problems

Problems involving tickets or stamps are very much like coin problems. Each type of ticket and stamp has a value, just like each type of coin does. So to solve these problems, we will follow the same steps we used to solve coin problems.

## Example 2.40

Danny paid $15.75 for stamps. The number of 49-cent stamps was five less than three times the number of 35-cent stamps. How many 49-cent stamps and how many 35-cent stamps did Danny buy?

## Try It 2.79

Eric paid $19.88 for stamps. The number of 49-cent stamps was eight more than twice the number of 35-cent stamps. How many 49-cent stamps and how many 35-cent stamps did Eric buy?

## Try It 2.80

Kailee paid $14.74 for stamps. The number of 49-cent stamps was four less than three times the number of 20-cent stamps. How many 49-cent stamps and how many 20-cent stamps did Kailee buy?

In most of our examples so far, we have been told that one quantity is four more than twice the other, or something similar. In our next example, we have to relate the quantities in a different way.

Suppose Aniket sold a total of 100 tickets. Each ticket was either an adult ticket or a child ticket. If he sold 20 child tickets, how many adult tickets did he sell?

Did you say “80”? How did you figure that out? Did you subtract 20 from 100?

If he sold 45 child tickets, how many adult tickets did he sell?

Did you say “55”? How did you find it? By subtracting 45 from 100?

Now, suppose Aniket sold x child tickets. Then how many adult tickets did he sell? To find out, we would follow the same logic we used above. In each case, we subtracted the number of child tickets from 100 to get the number of adult tickets. We now do the same with x .

We have summarized this in the table.

We will apply this technique in the next example.

## Example 2.41

A whale-watching ship had 40 paying passengers on board. The total revenue collected from tickets was $1,196. Full-fare passengers paid $32 each and reduced-fare passengers paid $26 each. How many full-fare passengers and how many reduced-fare passengers were on the ship?

## Try It 2.81

During her shift at the museum ticket booth, Leah sold 115 tickets for a total of $1,163. Adult tickets cost $12 and student tickets cost $5. How many adult tickets and how many student tickets did Leah sell?

## Try It 2.82

Galen sold 810 tickets for his church’s carnival for a total revenue of $2,820. Children’s tickets cost $3 each and adult tickets cost $5 each. How many children’s tickets and how many adult tickets did he sell?

Solve Mixture Word Problems

Now we’ll solve some more general applications of the mixture model. In mixture problems, we are often mixing two quantities, such as raisins and nuts, to create a mixture, such as trail mix. In our tables we will have a row for each item to be mixed as well as one for the final mixture.

## Example 2.42

Henning is mixing raisins and nuts to make 25 pounds of trail mix. Raisins cost $4.50 a pound and nuts cost $8 a pound. If Henning wants his cost for the trail mix to be $6.60 a pound, how many pounds of raisins and how many pounds of nuts should he use?

## Try It 2.83

Orlando is mixing nuts and cereal squares to make a party mix. Nuts sell for $7 a pound and cereal squares sell for $4 a pound. Orlando wants to make 30 pounds of party mix at a cost of $6.50 a pound, how many pounds of nuts and how many pounds of cereal squares should he use?

## Try It 2.84

Becca wants to mix fruit juice and soda to make a punch. She can buy fruit juice for $3 a gallon and soda for $4 a gallon. If she wants to make 28 gallons of punch at a cost of $3.25 a gallon, how many gallons of fruit juice and how many gallons of soda should she buy?

Solve Uniform Motion Applications

When you are driving down the interstate using your cruise control, the speed of your car stays the same—it is uniform. We call a problem in which the speed of an object is constant a uniform motion application. We will use the distance, rate, and time formula, D = r t , D = r t , to compare two scenarios, such as two vehicles travelling at different rates or in opposite directions.

Our problem solving strategies will still apply here, but we will add to the first step. The first step will include drawing a diagram that shows what is happening in the example. Drawing the diagram helps us understand what is happening so that we will write an appropriate equation. Then we will make a table to organize the information, like we did for the coin, ticket, and stamp applications.

The steps are listed here for easy reference:

## Solve a uniform motion application.

- Draw a diagram to illustrate what is happening.
- Label the columns rate, time, distance.
- List the two scenarios.
- Write in the information you know.
- Complete the chart.
- Use variable expressions to represent that quantity in each row.
- Multiply the rate times the time to get the distance.
- Restate the problem in one sentence with all the important information.
- Then, translate the sentence into an equation.

## Example 2.43

Wayne and Dennis like to ride the bike path from Riverside Park to the beach. Dennis’s speed is seven miles per hour faster than Wayne’s speed, so it takes Wayne two hours to ride to the beach while it takes Dennis 1.5 hours for the ride. Find the speed of both bikers.

Step 1. Read the problem. Make sure all the words and ideas are understood.

- Label the columns “Rate,” “Time,” and “Distance.”

You are asked to find the speed of both bikers.

Notice that the distance formula uses the word “rate,” but it is more common to use “speed”

when we talk about vehicles in everyday English.

Step 3. Name what we are looking for. Choose a variable to represent that quantity.

- Complete the chart

Step 4. Translate into an equation.

Step 5. Solve the equation using algebra techniques.

Dennis 28 mph ( 1.5 hours ) = 42 miles Wayne 21 mph ( 2 hours ) = 42 miles ✓ Dennis 28 mph ( 1.5 hours ) = 42 miles Wayne 21 mph ( 2 hours ) = 42 miles ✓

Wayne rode at 21 mph and Dennis rode at 28 mph.

## Try It 2.85

An express train and a local train leave Pittsburgh to travel to Washington, D.C. The express train can make the trip in four hours and the local train takes five hours for the trip. The speed of the express train is 12 miles per hour faster than the speed of the local train. Find the speed of both trains.

## Try It 2.86

Jeromy can drive from his house in Cleveland to his college in Chicago in 4.5 hours. It takes his mother six hours to make the same drive. Jeromy drives 20 miles per hour faster than his mother. Find Jeromy’s speed and his mother’s speed.

In Example 2.43 , we had two bikers traveling the same distance. In the next example, two people drive toward each other until they meet.

## Example 2.44

Carina is driving from her home in Anaheim to Berkeley on the same day her brother is driving from Berkeley to Anaheim, so they decide to meet for lunch along the way in Buttonwillow. The distance from Anaheim to Berkeley is 395 miles. It takes Carina three hours to get to Buttonwillow, while her brother drives four hours to get there. Carina’s average speed is 15 miles per hour faster than her brother’s average speed. Find Carina’s and her brother’s average speeds.

Step 2. Identify what we are looking for.

We are asked to find the average speeds of Carina and her brother.

- Use variable expressions to represent that quantity in each row. We are looking for their average speeds. Let’s let r represent the average speed of Carina's brother. Since Carina’s speed is 15 mph faster, we represent that as r + 15 . r + 15 . Fill in the speeds into the chart.

Step 6. Check the answer in the problem and make sure it makes sense. Carina drove 65 mph ( 3 hours ) = 195 miles Her brother drove 50 mph ( 4 hours ) = 200 miles ————— 395 miles ✓ Carina drove 65 mph ( 3 hours ) = 195 miles Her brother drove 50 mph ( 4 hours ) = 200 miles ————— 395 miles ✓

Carina drove 65 mph and her brother 50 mph.

## Try It 2.87

Christopher and his parents live 115 miles apart. They met at a restaurant between their homes to celebrate his mother’s birthday. Christopher drove one and a half hours while his parents drove one hour to get to the restaurant. Christopher’s average speed was ten miles per hour faster than his parents’ average speed. What were the average speeds of Christopher and of his parents as they drove to the restaurant?

## Try It 2.88

Ashley goes to college in Minneapolis, 234 miles from her home in Sioux Falls. She wants her parents to bring her more winter clothes, so they decide to meet at a restaurant on the road between Minneapolis and Sioux Falls. Ashley and her parents both drove two hours to the restaurant. Ashley’s average speed was seven miles per hour faster than her parents’ average speed. Find Ashley’s and her parents’ average speed.

As you read the next example, think about the relationship of the distances traveled. Which of the previous two examples is more similar to this situation?

## Example 2.45

Two truck drivers leave a rest area on the interstate at the same time. One truck travels east and the other one travels west. The truck traveling west travels at 70 mph and the truck traveling east has an average speed of 60 mph. How long will they travel before they are 325 miles apart?

Step 1. Read the problem. Make all the words and ideas are understood.

We are asked to find the amount of time the trucks will travel until they are 325 miles apart.

- Use variable expressions to represent that quantity in each row. We are looking for the time travelled. Both trucks will travel the same amount of time. Let’s call the time t . Since their speeds are different, they will travel different distances.

Now solve this equation 70 t + 60 t = 325 130 t = 325 t = 2.5 Now solve this equation 70 t + 60 t = 325 130 t = 325 t = 2.5

So it will take the trucks 2.5 2.5 hours to be 325 miles apart.

Truck going West 70 mph ( 2.5 hours ) = 175 miles Truck going East 60 mph ( 2.5 hours ) = 150 miles ————— 325 miles ✓ Truck going West 70 mph ( 2.5 hours ) = 175 miles Truck going East 60 mph ( 2.5 hours ) = 150 miles ————— 325 miles ✓

Step 7. Answer the question with a complete sentence. It will take the trucks 2.5 hours to be 325 miles apart.

## Try It 2.89

Pierre and Monique leave their home in Portland at the same time. Pierre drives north on the turnpike at a speed of 75 miles per hour while Monique drives south at a speed of 68 miles per hour. How long will it take them to be 429 miles apart?

## Try It 2.90

Thanh and Nhat leave their office in Sacramento at the same time. Thanh drives north on I-5 at a speed of 72 miles per hour. Nhat drives south on I-5 at a speed of 76 miles per hour. How long will it take them to be 330 miles apart?

It is important to make sure that the units match when we use the distance rate and time formula. For instance, if the rate is in miles per hour, then the time must be in hours.

## Example 2.46

When Naoko walks to school, it takes her 30 minutes. If she rides her bike, it takes her 15 minutes. Her speed is three miles per hour faster when she rides her bike than when she walks. What is her speed walking and her speed riding her bike?

First, we draw a diagram that represents the situation to help us see what is happening.

We are asked to find her speed walking and riding her bike. Let’s call her walking speed r . Since her biking speed is three miles per hour faster, we will call that speed r + 3 . r + 3 . We write the speeds in the chart.

The speed is in miles per hour, so we need to express the times in hours, too, in order for the units to be the same. Remember, 1 hour is 60 minutes. So:

We write the times in the chart.

Next, we multiply rate times time to fill in the distance column.

The equation will come from the fact that the distance from Naoko’s home to her school is the same whether she is walking or riding her bike. So we say:

Yes, either way Naoko travels 1.5 miles to school.

Naoko’s walking speed is 3 mph and her speed riding her bike is 6 mph.

## Try It 2.91

Suzy takes 50 minutes to hike uphill from the parking lot to the lookout tower. It takes her 30 minutes to hike back down to the parking lot. Her speed going downhill is 1.2 miles per hour faster than her speed going uphill. Find Suzy’s uphill and downhill speeds.

## Try It 2.92

Llewyn takes 45 minutes to drive his boat upstream from the dock to his favorite fishing spot. It takes him 30 minutes to drive the boat back downstream to the dock. The boat’s speed going downstream is four miles per hour faster than its speed going upstream. Find the boat’s upstream and downstream speeds.

In the distance, rate and time formula, time represents the actual amount of elapsed time (in hours, minutes, etc.). If a problem gives us starting and ending times as clock times, we must find the elapsed time in order to use the formula.

## Example 2.47

Cruz is training to compete in a triathlon. He left his house at 6:00 and ran until 7:30. Then he rode his bike until 9:45. He covered a total distance of 51 miles. His speed when biking was 1.6 times his speed when running. Find Cruz’s biking and running speeds.

A diagram will help us model this trip.

Next, we create a table to organize the information. We know the total distance is 51 miles. We are looking for the rate of speed for each part of the trip. The rate while biking is 1.6 times the rate of running. If we let r = the rate running, then the rate biking is 1.6 r .

The times here are given as clock times. Cruz started from home at 6:00 a.m. and started biking at 7:30 a.m. So he spent 1.5 hours running. Then he biked from 7:30 a.m until 9:45 a.m. So he spent 2.25 hours biking.

Now, we multiply the rates by the times.

By looking at the diagram, we can see that the sum of the distance running and the distance biking is 255 miles.

## Try It 2.93

Hamilton loves to travel to Las Vegas, 255 miles from his home in Orange County. On his last trip, he left his house at 2:00 p.m. The first part of his trip was on congested city freeways. At 4:00 pm, the traffic cleared and he was able to drive through the desert at a speed 1.75 times as fast as when he drove in the congested area. He arrived in Las Vegas at 6:30 p.m. How fast was he driving during each part of his trip?

## Try It 2.94

Phuong left home on his bicycle at 10:00. He rode on the flat street until 11:15, then rode uphill until 11:45. He rode a total of 31 miles. His speed riding uphill was 0.6 times his speed on the flat street. Find his speed biking uphill and on the flat street.

## Practice Makes Perfect

In the following exercises, solve each coin word problem.

Michaela has $2.05 in dimes and nickels in her change purse. She has seven more dimes than nickels. How many coins of each type does she have?

Liliana has $2.10 in nickels and quarters in her backpack. She has 12 more nickels than quarters. How many coins of each type does she have?

In a cash drawer there is $125 in $5 and $10 bills. The number of $10 bills is twice the number of $5 bills. How many of each type of bill is in the drawer?

Sumanta has $175 in $5 and $10 bills in his drawer. The number of $5 bills is three times the number of $10 bills. How many of each are in the drawer?

Chi has $11.30 in dimes and quarters. The number of dimes is three more than three times the number of quarters. How many of each are there?

Alison has $9.70 in dimes and quarters. The number of quarters is eight more than four times the number of dimes. How many of each coin does she have?

Mukul has $3.75 in quarters, dimes and nickels in his pocket. He has five more dimes than quarters and nine more nickels than quarters. How many of each coin are in his pocket?

Vina has $4.70 in quarters, dimes and nickels in her purse. She has eight more dimes than quarters and six more nickels than quarters. How many of each coin are in her purse?

In the following exercises, solve each ticket or stamp word problem.

The first day of a water polo tournament the total value of tickets sold was $17,610. One-day passes sold for $20 and tournament passes sold for $30. The number of tournament passes sold was 37 more than the number of day passes sold. How many day passes and how many tournament passes were sold?

At the movie theater, the total value of tickets sold was $2,612.50. Adult tickets sold for $10 each and senior/child tickets sold for $7.50 each. The number of senior/child tickets sold was 25 less than twice the number of adult tickets sold. How many senior/child tickets and how many adult tickets were sold?

Julie went to the post office and bought both $0.41 stamps and $0.26 postcards. She spent $51.40. The number of stamps was 20 more than twice the number of postcards. How many of each did she buy?

Jason went to the post office and bought both $0.41 stamps and $0.26 postcards and spent $10.28 The number of stamps was four more than twice the number of postcards. How many of each did he buy?

Hilda has $210 worth of $10 and $12 stock shares. The number of $10 shares is five more than twice the number of $12 shares. How many of each type of share does she have?

Mario invested $475 in $45 and $25 stock shares. The number of $25 shares was five less than three times the number of $45 shares. How many of each type of share did he buy?

The ice rink sold 95 tickets for the afternoon skating session, for a total of $828. General admission tickets cost $10 each and youth tickets cost $8 each. How many general admission tickets and how many youth tickets were sold?

For the 7:30 show time, 140 movie tickets were sold. Receipts from the $13 adult tickets and the $10 senior tickets totaled $1,664. How many adult tickets and how many senior tickets were sold?

The box office sold 360 tickets to a concert at the college. The total receipts were $4,170. General admission tickets cost $15 and student tickets cost $10. How many of each kind of ticket was sold?

Last Saturday, the museum box office sold 281 tickets for a total of $3,954. Adult tickets cost $15 and student tickets cost $12. How many of each kind of ticket was sold?

In the following exercises, solve each mixture word problem.

Macario is making 12 pounds of nut mixture with macadamia nuts and almonds. Macadamia nuts cost $9 per pound and almonds cost $5.25 per pound. How many pounds of macadamia nuts and how many pounds of almonds should Macario use for the mixture to cost $6.50 per pound to make?

Carmen wants to tile the floor of his house. He will need 1,000 square feet of tile. He will do most of the floor with a tile that costs $1.50 per square foot, but also wants to use an accent tile that costs $9.00 per square foot. How many square feet of each tile should he plan to use if he wants the overall cost to be $3 per square foot?

Riley is planning to plant a lawn in his yard. He will need nine pounds of grass seed. He wants to mix Bermuda seed that costs $4.80 per pound with Fescue seed that costs $3.50 per pound. How much of each seed should he buy so that the overall cost will be $4.02 per pound?

Vartan was paid $25,000 for a cell phone app that he wrote and wants to invest it to save for his son’s education. He wants to put some of the money into a bond that pays 4% annual interest and the rest into stocks that pay 9% annual interest. If he wants to earn 7.4% annual interest on the total amount, how much money should he invest in each account?

Vern sold his 1964 Ford Mustang for $55,000 and wants to invest the money to earn him 5.8% interest per year. He will put some of the money into Fund A that earns 3% per year and the rest in Fund B that earns 10% per year. How much should he invest into each fund if he wants to earn 5.8% interest per year on the total amount?

Dominic pays 7% interest on his $15,000 college loan and 12% interest on his $11,000 car loan. What average interest rate does he pay on the total $26,000 he owes? (Round your answer to the nearest tenth of a percent.)

Liam borrowed a total of $35,000 to pay for college. He pays his parents 3% interest on the $8,000 he borrowed from them and pays the bank 6.8% on the rest. What average interest rate does he pay on the total $35,000? (Round your answer to the nearest tenth of a percent.)

In the following exercises, solve.

Lilah is moving from Portland to Seattle. It takes her three hours to go by train. Mason leaves the train station in Portland and drives to the train station in Seattle with all Lilah’s boxes in his car. It takes him 2.4 hours to get to Seattle, driving at 15 miles per hour faster than the speed of the train. Find Mason’s speed and the speed of the train.

Kathy and Cheryl are walking in a fundraiser. Kathy completes the course in 4.8 hours and Cheryl completes the course in eight hours. Kathy walks two miles per hour faster than Cheryl. Find Kathy’s speed and Cheryl’s speed.

Two busses go from Sacramento to San Diego. The express bus makes the trip in 6.8 hours and the local bus takes 10.2 hours for the trip. The speed of the express bus is 25 mph faster than the speed of the local bus. Find the speed of both busses.

A commercial jet and a private airplane fly from Denver to Phoenix. It takes the commercial jet 1.6 hours for the flight, and it takes the private airplane 2.6 hours. The speed of the commercial jet is 210 miles per hour faster than the speed of the private airplane. Find the speed of both airplanes to the nearest 10 mph.

Saul drove his truck three hours from Dallas towards Kansas City and stopped at a truck stop to get dinner. At the truck stop he met Erwin, who had driven four hours from Kansas City towards Dallas. The distance between Dallas and Kansas City is 542 miles, and Erwin’s speed was eight miles per hour slower than Saul’s speed. Find the speed of the two truckers.

Charlie and Violet met for lunch at a restaurant between Memphis and New Orleans. Charlie had left Memphis and drove 4.8 hours towards New Orleans. Violet had left New Orleans and drove two hours towards Memphis, at a speed 10 miles per hour faster than Charlie’s speed. The distance between Memphis and New Orleans is 394 miles. Find the speed of the two drivers.

Sisters Helen and Anne live 332 miles apart. For Thanksgiving, they met at their other sister’s house partway between their homes. Helen drove 3.2 hours and Anne drove 2.8 hours. Helen’s average speed was four miles per hour faster than Anne’s. Find Helen’s average speed and Anne’s average speed.

Ethan and Leo start riding their bikes at the opposite ends of a 65-mile bike path. After Ethan has ridden 1.5 hours and Leo has ridden two hours, they meet on the path. Ethan’s speed is six miles per hour faster than Leo’s speed. Find the speed of the two bikers.

Elvira and Aletheia live 3.1 miles apart on the same street. They are in a study group that meets at a coffee shop between their houses. It took Elvira half an hour and Aletheia two-thirds of an hour to walk to the coffee shop. Aletheia’s speed is 0.6 miles per hour slower than Elvira’s speed. Find both women’s walking speeds.

DaMarcus and Fabian live 23 miles apart and play soccer at a park between their homes. DaMarcus rode his bike for three-quarters of an hour and Fabian rode his bike for half an hour to get to the park. Fabian’s speed was six miles per hour faster than DaMarcus’ speed. Find the speed of both soccer players.

Cindy and Richard leave their dorm in Charleston at the same time. Cindy rides her bicycle north at a speed of 18 miles per hour. Richard rides his bicycle south at a speed of 14 miles per hour. How long will it take them to be 96 miles apart?

Matt and Chris leave their uncle’s house in Phoenix at the same time. Matt drives west on I-60 at a speed of 76 miles per hour. Chris drives east on I-60 at a speed of 82 miles per hour. How many hours will it take them to be 632 miles apart?

Two busses leave Billings at the same time. The Seattle bus heads west on I-90 at a speed of 73 miles per hour while the Chicago bus heads east at a speed of 79 miles an hour. How many hours will it take them to be 532 miles apart?

Two boats leave the same dock in Cairo at the same time. One heads north on the Mississippi River while the other heads south. The northbound boat travels four miles per hour. The southbound boat goes eight miles per hour. How long will it take them to be 54 miles apart?

Lorena walks the path around the park in 30 minutes. If she jogs, it takes her 20 minutes. Her jogging speed is 1.5 miles per hour faster than her walking speed. Find Lorena’s walking speed and jogging speed.

Julian rides his bike uphill for 45 minutes, then turns around and rides back downhill. It takes him 15 minutes to get back to where he started. His uphill speed is 3.2 miles per hour slower than his downhill speed. Find Julian’s uphill and downhill speed.

Cassius drives his boat upstream for 45 minutes. It takes him 30 minutes to return downstream. His speed going upstream is three miles per hour slower than his speed going downstream. Find his upstream and downstream speeds.

It takes Darline 20 minutes to drive to work in light traffic. To come home, when there is heavy traffic, it takes her 36 minutes. Her speed in light traffic is 24 miles per hour faster than her speed in heavy traffic. Find her speed in light traffic and in heavy traffic.

At 1:30, Marlon left his house to go to the beach, a distance of 7.6 miles. He rode his skateboard until 2:15, and then walked the rest of the way. He arrived at the beach at 3:00. Marlon’s speed on his skateboard is 2.5 times his walking speed. Find his speed when skateboarding and when walking.

Aaron left at 9:15 to drive to his mountain cabin 108 miles away. He drove on the freeway until 10:45 and then drove on a mountain road. He arrived at 11:05. His speed on the freeway was three times his speed on the mountain road. Find Aaron’s speed on the freeway and on the mountain road.

Marisol left Los Angeles at 2:30 to drive to Santa Barbara, a distance of 95 miles. The traffic was heavy until 3:20. She drove the rest of the way in very light traffic and arrived at 4:20. Her speed in heavy traffic was 40 miles per hour slower than her speed in light traffic. Find her speed in heavy traffic and in light traffic.

Lizette is training for a marathon. At 7:00 she left her house and ran until 8:15 then she walked until 11:15. She covered a total distance of 19 miles. Her running speed was five miles per hour faster than her walking speed. Find her running and walking speeds.

## Everyday Math

John left his house in Irvine at 8:35 a.m. to drive to a meeting in Los Angeles, 45 miles away. He arrived at the meeting at 9:50 a.m.. At 5:30 p.m. he left the meeting and drove home. He arrived home at 7:18 p.m.

ⓐ What was his average speed on the drive from Irvine to Los Angeles?

ⓑ What was his average speed on the drive from Los Angeles to Irvine?

ⓒ What was the total time he spent driving to and from this meeting?

Sarah wants to arrive at her friend’s wedding at 3:00. The distance from Sarah’s house to the wedding is 95 miles. Based on usual traffic patterns, Sarah predicts she can drive the first 15 miles at 60 miles per hour, the next 10 miles at 30 miles per hour, and the remainder of the drive at 70 miles per hour.

ⓐ How long will it take Sarah to drive the first 15 miles?

ⓑ How long will it take Sarah to drive the next 10 miles?

ⓒ How long will it take Sarah to drive the rest of the trip?

ⓓ What time should Sarah leave her house?

## Writing Exercises

Suppose you have six quarters, nine dimes, and four pennies. Explain how you find the total value of all the coins.

Do you find it helpful to use a table when solving coin problems? Why or why not?

In the table used to solve coin problems, one column is labeled “number” and another column is labeled “value.” What is the difference between the “number” and the “value”?

When solving a uniform motion problem, how does drawing a diagram of the situation help you?

ⓐ After completing the exercises, use this checklist to evaluate your mastery of the objectives of this section.

ⓑ On a scale of 1-10, how would you rate your mastery of this section in light of your responses on the checklist? How can you improve this?

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Chapter 8: Rational Expressions

## 8.8 Rate Word Problems: Speed, Distance and Time

Distance, rate and time problems are a standard application of linear equations. When solving these problems, use the relationship rate (speed or velocity) times time equals distance .

[latex]r\cdot t=d[/latex]

For example, suppose a person were to travel 30 km/h for 4 h. To find the total distance, multiply rate times time or (30km/h)(4h) = 120 km.

The problems to be solved here will have a few more steps than described above. So to keep the information in the problem organized, use a table. An example of the basic structure of the table is below:

The third column, distance, will always be filled in by multiplying the rate and time columns together. If given a total distance of both persons or trips, put this information in the distance column. Now use this table to set up and solve the following examples.

Example 8.8.1

Joey and Natasha start from the same point and walk in opposite directions. Joey walks 2 km/h faster than Natasha. After 3 hours, they are 30 kilometres apart. How fast did each walk?

The distance travelled by both is 30 km. Therefore, the equation to be solved is:

[latex]\begin{array}{rrrrrrl} 3r&+&3(r&+&2)&=&30 \\ 3r&+&3r&+&6&=&30 \\ &&&-&6&&-6 \\ \hline &&&&\dfrac{6r}{6}&=&\dfrac{24}{6} \\ \\ &&&&r&=&4 \text{ km/h} \end{array}[/latex]

This means that Natasha walks at 4 km/h and Joey walks at 6 km/h.

Example 8.8.2

Nick and Chloe left their campsite by canoe and paddled downstream at an average speed of 12 km/h. They turned around and paddled back upstream at an average rate of 4 km/h. The total trip took 1 hour. After how much time did the campers turn around downstream?

The distance travelled downstream is the same distance that they travelled upstream. Therefore, the equation to be solved is:

[latex]\begin{array}{rrlll} 12(t)&=&4(1&-&t) \\ 12t&=&4&-&4t \\ +4t&&&+&4t \\ \hline \dfrac{16t}{16}&=&\dfrac{4}{16}&& \\ \\ t&=&0.25&& \end{array}[/latex]

This means the campers paddled downstream for 0.25 h and spent 0.75 h paddling back.

Example 8.8.3

Terry leaves his house riding a bike at 20 km/h. Sally leaves 6 h later on a scooter to catch up with him travelling at 80 km/h. How long will it take her to catch up with him?

The distance travelled by both is the same. Therefore, the equation to be solved is:

[latex]\begin{array}{rrrrr} 20(t)&=&80(t&-&6) \\ 20t&=&80t&-&480 \\ -80t&&-80t&& \\ \hline \dfrac{-60t}{-60}&=&\dfrac{-480}{-60}&& \\ \\ t&=&8&& \end{array}[/latex]

This means that Terry travels for 8 h and Sally only needs 2 h to catch up to him.

Example 8.8.4

On a 130-kilometre trip, a car travelled at an average speed of 55 km/h and then reduced its speed to 40 km/h for the remainder of the trip. The trip took 2.5 h. For how long did the car travel 40 km/h?

[latex]\begin{array}{rrrrrrr} 55(t)&+&40(2.5&-&t)&=&130 \\ 55t&+&100&-&40t&=&130 \\ &-&100&&&&-100 \\ \hline &&&&\dfrac{15t}{15}&=&\dfrac{30}{15} \\ \\ &&&&t&=&2 \end{array}[/latex]

This means that the time spent travelling at 40 km/h was 0.5 h.

Distance, time and rate problems have a few variations that mix the unknowns between distance, rate and time. They generally involve solving a problem that uses the combined distance travelled to equal some distance or a problem in which the distances travelled by both parties is the same. These distance, rate and time problems will be revisited later on in this textbook where quadratic solutions are required to solve them.

For Questions 1 to 8, find the equations needed to solve the problems. Do not solve.

- A is 60 kilometres from B. An automobile at A starts for B at the rate of 20 km/h at the same time that an automobile at B starts for A at the rate of 25 km/h. How long will it be before the automobiles meet?
- Two automobiles are 276 kilometres apart and start to travel toward each other at the same time. They travel at rates differing by 5 km/h. If they meet after 6 h, find the rate of each.
- Two trains starting at the same station head in opposite directions. They travel at the rates of 25 and 40 km/h, respectively. If they start at the same time, how soon will they be 195 kilometres apart?
- Two bike messengers, Jerry and Susan, ride in opposite directions. If Jerry rides at the rate of 20 km/h, at what rate must Susan ride if they are 150 kilometres apart in 5 hours?
- A passenger and a freight train start toward each other at the same time from two points 300 kilometres apart. If the rate of the passenger train exceeds the rate of the freight train by 15 km/h, and they meet after 4 hours, what must the rate of each be?
- Two automobiles started travelling in opposite directions at the same time from the same point. Their rates were 25 and 35 km/h, respectively. After how many hours were they 180 kilometres apart?
- A man having ten hours at his disposal made an excursion by bike, riding out at the rate of 10 km/h and returning on foot at the rate of 3 km/h. Find the distance he rode.
- A man walks at the rate of 4 km/h. How far can he walk into the country and ride back on a trolley that travels at the rate of 20 km/h, if he must be back home 3 hours from the time he started?

Solve Questions 9 to 22.

- A boy rides away from home in an automobile at the rate of 28 km/h and walks back at the rate of 4 km/h. The round trip requires 2 hours. How far does he ride?
- A motorboat leaves a harbour and travels at an average speed of 15 km/h toward an island. The average speed on the return trip was 10 km/h. How far was the island from the harbour if the trip took a total of 5 hours?
- A family drove to a resort at an average speed of 30 km/h and later returned over the same road at an average speed of 50 km/h. Find the distance to the resort if the total driving time was 8 hours.
- As part of his flight training, a student pilot was required to fly to an airport and then return. The average speed to the airport was 90 km/h, and the average speed returning was 120 km/h. Find the distance between the two airports if the total flying time was 7 hours.
- Sam starts travelling at 4 km/h from a campsite 2 hours ahead of Sue, who travels 6 km/h in the same direction. How many hours will it take for Sue to catch up to Sam?
- A man travels 5 km/h. After travelling for 6 hours, another man starts at the same place as the first man did, following at the rate of 8 km/h. When will the second man overtake the first?
- A motorboat leaves a harbour and travels at an average speed of 8 km/h toward a small island. Two hours later, a cabin cruiser leaves the same harbour and travels at an average speed of 16 km/h toward the same island. How many hours after the cabin cruiser leaves will it be alongside the motorboat?
- A long distance runner started on a course, running at an average speed of 6 km/h. One hour later, a second runner began the same course at an average speed of 8 km/h. How long after the second runner started will they overtake the first runner?
- Two men are travelling in opposite directions at the rate of 20 and 30 km/h at the same time and from the same place. In how many hours will they be 300 kilometres apart?
- Two trains start at the same time from the same place and travel in opposite directions. If the rate of one is 6 km/h more than the rate of the other and they are 168 kilometres apart at the end of 4 hours, what is the rate of each?
- Two cyclists start from the same point and ride in opposite directions. One cyclist rides twice as fast as the other. In three hours, they are 72 kilometres apart. Find the rate of each cyclist.
- Two small planes start from the same point and fly in opposite directions. The first plane is flying 25 km/h slower than the second plane. In two hours, the planes are 430 kilometres apart. Find the rate of each plane.
- On a 130-kilometre trip, a car travelled at an average speed of 55 km/h and then reduced its speed to 40 km/h for the remainder of the trip. The trip took a total of 2.5 hours. For how long did the car travel at 40 km/h?
- Running at an average rate of 8 m/s, a sprinter ran to the end of a track and then jogged back to the starting point at an average of 3 m/s. The sprinter took 55 s to run to the end of the track and jog back. Find the length of the track.

Answer Key 8.8

Intermediate Algebra by Terrance Berg is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International License , except where otherwise noted.

## Share This Book

## Speed and Velocity

Speed is how fast something moves.

Velocity is speed with a direction .

Saying Ariel the Dog runs at 9 km/h (kilometers per hour) is a speed.

But saying he runs 9 km/h Westwards is a velocity.

Imagine something moving back and forth very fast: it has a high speed, but a low (or zero) velocity.

Speed is measured as distance moved over time.

Speed = Distance Time

## Example: A car travels 50 km in one hour.

Its average speed is 50 km per hour (50 km/h)

Speed = Distance Time = 50 km 1 hour

We can also use these symbols:

Speed = Δs Δt

Where Δ (" Delta ") means "change in", and

- s means distance ("s" for "space")
- t means time

## Example: You run 360 m in 60 seconds.

So your speed is 6 meters per second (6 m/s).

Speed is commonly measured in:

- meters per second (m/s or m s -1 ), or
- kilometers per hour (km/h or km h -1 )

A km is 1000 m, and there are 3600 seconds in an hour, so we can convert like this (see Unit Conversion Method to learn more):

1 m 1 s × 1 km 1000 m × 3600 s 1 h = 3600 m · km · s 1000 s · m · h = 3.6 km 1 h

So 1 m/s is equal to 3.6 km/h

## Example: What is 20 m/s in km/h ?

20 m/s × 3.6 km/h 1 m/s = 72 km/h

## Example: What is 120 km/h in m/s ?

120 km/h × 1 m/s 3.6 km/h = 33.333... m/s

## Average vs Instantaneous Speed

The examples so far calculate average speed : how far something travels over a period of time.

But speed can change as time goes by. A car can go faster and slower, maybe even stop at lights.

So there is also instantaneous speed : the speed at an instant in time. We can try to measure it by using a very short span of time (the shorter the better).

## Example: Sam uses a stopwatch and measures 1.6 seconds as the car travels between two posts 20 m apart. What is the instantaneous speed ?

Well, we don't know exactly, as the car may have been speeding up or slowing down during that time, but we can estimate:

20 m 1.6 s = 12.5 m/s = 45 km/h

It is really still an average, but is close to an instantaneous speed.

## Constant Speed

When the speed does not change it is constant .

For constant speed, the average and instantaneous speeds are the same.

Because the direction is important velocity uses displacement instead of distance:

Velocity = Displacement Time in a direction.

## Example: You walk from home to the shop in 100 seconds, what is your speed and what is your velocity?

Speed = 220 m 100 s = 2.2 m/s

Velocity = 130 m 100 s East = 1.3 m/s East

## You forgot your money so you turn around and go back home in 120 more seconds: what is your round-trip speed and velocity?

The total time is 100 s + 120 s = 220 s:

Speed = 440 m 220 s = 2.0 m/s

Velocity = 0 m 220 s = 0 m/s

Yes, the velocity is zero as you ended up where you started.

Learn more at Vectors .

Motion is relative. When we say something is "at rest" or "moving at 4 m/s" we forget to say "in relation to me" or "in relation to the ground", etc.

Think about this: are you really standing still? You are on planet Earth which is spinning at 40,075 km per day (about 1675 km/h or 465 m/s), and moving around the Sun at about 100,000 km/h, which is itself moving through the Galaxy.

Next time you are out walking, imagine you are still and it is the world that moves under your feet. Feels great.

It is all relative!

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## 3.5: Solve Uniform Motion Applications

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## Learning Objectives

By the end of this section, you will be able to:

- Solve uniform motion applications

Before you get started, take this readiness quiz.

- Find the distance traveled by a car going 70 miles per hour for 3 hours. If you missed this problem, review Exercise 2.6.1 .
- Solve \(x+1.2(x−10)=98\). If you missed this problem, review Exercise 2.4.7 .
- Convert 90 minutes to hours. If you missed this problem, review Exercise 1.11.1 .

## Solve Uniform Motion Applications

When planning a road trip, it often helps to know how long it will take to reach the destination or how far to travel each day. We would use the distance, rate, and time formula, D=rt, which we have already seen.

In this section, we will use this formula in situations that require a little more algebra to solve than the ones we saw earlier. Generally, we will be looking at comparing two scenarios, such as two vehicles traveling at different rates or in opposite directions. When the speed of each vehicle is constant, we call applications like this uniform motion problems .

Our problem-solving strategies will still apply here, but we will add to the first step. The first step will include drawing a diagram that shows what is happening in the example. Drawing the diagram helps us understand what is happening so that we will write an appropriate equation. Then we will make a table to organize the information, like we did for the money applications.

The steps are listed here for easy reference:

## USE A PROBLEM-SOLVING STRATEGY IN DISTANCE, RATE, AND TIME APPLICATIONS.

Draw a diagram to illustrate what it happening.

Create a table to organize the information.

Label the columns rate, time, distance.

List the two scenarios.

Write in the information you know.

- Identify what we are looking for.
- Complete the chart.
- Use variable expressions to represent that quantity in each row.

Multiply the rate times the time to get the distance.

- Restate the problem in one sentence with all the important information.
- Then, translate the sentence into an equation.
- Solve the equation using good algebra techniques.
- Check the answer in the problem and make sure it makes sense.
- Answer the question with a complete sentence.

## Example \(\PageIndex{1}\)

An express train and a local train leave Pittsburgh to travel to Washington, D.C. The express train can make the trip in 4 hours and the local train takes 5 hours for the trip. The speed of the express train is 12 miles per hour faster than the speed of the local train. Find the speed of both trains.

Step 1. Read the problem. Make sure all the words and ideas are understood.

Create a table to organize the information. Label the columns “Rate,” “Time,” and “Distance.” List the two scenarios. Write in the information you know.

Step 2. Identify what we are looking for.

We are asked to find the speed of both trains. Notice that the distance formula uses the word “rate,” but it is more common to use “speed” when we talk about vehicles in everyday English.

Step 3. Name what we are looking for. Choose a variable to represent that quantity.

Complete the chart Use variable expressions to represent that quantity in each row. We are looking for the speed of the trains. Let’s let r represent the speed of the local train. Since the speed of the express train is 12 mph faster, we represent that as r+12.

\[\begin{aligned} r &=\text { speed of the local train } \\ r+12 &=\text { speed of the express train } \end{aligned}\]

Fill in the speeds into the chart.

Step 4. Translate into an equation.

- The equation to model this situation will come from the relation between the distances. Look at the diagram we drew above. How is the distance traveled by the express train related to the distance traveled by the local train?
- Since both trains leave from Pittsburgh and travel to Washington, D.C. they travel the same distance. So we write:

Step 5. Solve the equation using good algebra techniques.

Step 7. Answer the question with a complete sentence.

## Try It \(\PageIndex{2}\)

Wayne and Dennis like to ride the bike path from Riverside Park to the beach. Dennis’s speed is seven miles per hour faster than Wayne’s speed, so it takes Wayne 2 hours to ride to the beach while it takes Dennis 1.5 hours for the ride. Find the speed of both bikers.

Wayne 21 mph, Dennis 28 mph

## Try It \(\PageIndex{3}\)

Jeromy can drive from his house in Cleveland to his college in Chicago in 4.5 hours. It takes his mother 6 hours to make the same drive. Jeromy drives 20 miles per hour faster than his mother. Find Jeromy’s speed and his mother’s speed.

Jeromy 80 mph, mother 60 mph

In Exercise \(\PageIndex{4}\), the last example, we had two trains traveling the same distance. The diagram and the chart helped us write the equation we solved. Let’s see how this works in another case.

## Example \(\PageIndex{4}\)

Christopher and his parents live 115 miles apart. They met at a restaurant between their homes to celebrate his mother’s birthday. Christopher drove 1.5 hours while his parents drove 1 hour to get to the restaurant. Christopher’s average speed was 10 miles per hour faster than his parents’ average speed. What were the average speeds of Christopher and of his parents as they drove to the restaurant?

We are asked to find the average speeds of Christopher and his parents.

The distance Christopher travelled plus the distance his parents travel must add up to 115 miles. So we write:

\(\begin{array} {cc} {} &{1.5(r + 10) + r = 115} \\ {} &{1.5r + 15 + r = 115} \\ {\text{Now solve this equation.}} &{2.5r + 15 = 115} \\{} &{2.5r = 100} \\{} &{r = 40} \\ {} &{\text{so the parents' speed was 40 mph.}} \\ {} &{r + 10} \\ {\text{Christopher's speed is r + 10}} &{40 + 10} \\ {} &{50} \\ {} &{\text{Christopher's speed was 50 mph.}} \\ {} &{} \end{array}\)

Step 6. Check the answer in the problem and make sure it makes sense.

\(\begin{array}{llll} {\text{Christopher drove}} &{50\text{ mph (1.5 hours)}} &{=} &{75\text{ miles}}\\ {\text{His parents drove}} &{40\text{ mph (1 hour)}} &{=} &{\underline{40 \text{ miles}}}\\ {} &{} &{} &{115\text{ miles}} \end{array}\)

\(\begin{array}{ll} {\textbf{Step 7. Answer}\text{ the question with a complete sentence.}} &{} \\{} &{\text{Christopher's speed was 50 mph.}}\\ {} &{\text{His parents' speed was 40 mph.}} \end{array}\)

## Try It \(\PageIndex{5}\)

Carina is driving from her home in Anaheim to Berkeley on the same day her brother is driving from Berkeley to Anaheim, so they decide to meet for lunch along the way in Buttonwillow. The distance from Anaheim to Berkeley is 410 miles. It takes Carina 3 hours to get to Buttonwillow, while her brother drives 4 hours to get there. The average speed Carina’s brother drove was 15 miles per hour faster than Carina’s average speed. Find Carina’s and her brother’s average speeds.

Carina 50 mph, brother 65 mph

## Try It \(\PageIndex{6}\)

Ashley goes to college in Minneapolis, 234 miles from her home in Sioux Falls. She wants her parents to bring her more winter clothes, so they decide to meet at a restaurant on the road between Minneapolis and Sioux Falls. Ashley and her parents both drove 2 hours to the restaurant. Ashley’s average speed was seven miles per hour faster than her parents’ average speed. Find Ashley’s and her parents’ average speed.

parents 55 mph, Ashley 62 mph

As you read the next example, think about the relationship of the distances traveled. Which of the previous two examples is more similar to this situation?

## Example \(\PageIndex{7}\)

Two truck drivers leave a rest area on the interstate at the same time. One truck travels east and the other one travels west. The truck traveling west travels at 70 mph and the truck traveling east has an average speed of 60 mph. How long will they travel before they are 325 miles apart?

We are asked to find the amount of time the trucks will travel until they are 325 miles apart.

We are looking for the time traveled. Both trucks will travel the same amount of time. Let’s call the time t . Since their speeds are different, they will travel different distances. Complete the chart.

We need to find a relation between the distances in order to write an equation. Looking at the diagram, what is the relationship between the distance each of the trucks will travel? The distance traveled by the truck going west plus the distance traveled by the truck going east must add up to 325 miles. So we write:

\[\begin{array} {lrll} {\text{Now solve this equation. }} & {70 t+60 t} &{=} &{325} \\ {} &{130 t} &{=} &{325} \\ {} &{t} &{=} &{2.5} \end{array}\]

\(\begin{array}{llll} {\text{Truck going West}} &{70\text{ mph (2.5 hours)}} &{=} &{175\text{ miles}}\\ {\text{Truck going East}} &{60\text{ mph (2.5 hour)}} &{=} &{\underline{150 \text{ miles}}}\\ {} &{} &{} &{325\text{ miles}} \end{array}\)

\(\begin{array}{ll} \\{\textbf{Step 7. Answer}\text{ the question with a complete sentence.}} &{\text{It will take the truck 2.5 hours to be 325 miles apart.}} \end{array}\)

## Try It \(\PageIndex{8}\)

Pierre and Monique leave their home in Portland at the same time. Pierre drives north on the turnpike at a speed of 75 miles per hour while Monique drives south at a speed of 68 miles per hour. How long will it take them to be 429 miles apart?

## Try It \(\PageIndex{9}\)

Thanh and Nhat leave their office in Sacramento at the same time. Thanh drives north on I-5 at a speed of 72 miles per hour. Nhat drives south on I-5 at a speed of 76 miles per hour. How long will it take them to be 330 miles apart?

## MATCHING UNITS IN PROBLEMS

It is important to make sure the units match when we use the distance rate and time formula. For instance, if the rate is in miles per hour, then the time must be in hours.

## Example \(\PageIndex{10}\)

When Katie Mae walks to school, it takes her 30 minutes. If she rides her bike, it takes her 15 minutes. Her speed is three miles per hour faster when she rides her bike than when she walks. What are her walking speed and her speed riding her bike?

First, we draw a diagram that represents the situation to help us see what is happening.

We are asked to find her speed walking and riding her bike. Let’s call her walking speed r . Since her biking speed is three miles per hour faster, we will call that speed r+3. We write the speeds in the chart.

The speed is in miles per hour, so we need to express the times in hours, too, in order for the units to be the same. Remember, one hour is 60 minutes. So:

\[\begin{array}{l}{30 \text { minutes is } \frac{30}{60} \text { or } \frac{1}{2} \text { hour }} \\ {15 \text { minutes is } \frac{15}{60} \text { or } \frac{1}{4} \text { hour }}\end{array}\]

Next, we multiply rate times time to fill in the distance column.

The equation will come from the fact that the distance from Katie Mae’s home to her school is the same whether she is walking or riding her bike.

## Try It \(\PageIndex{11}\)

Suzy takes 50 minutes to hike uphill from the parking lot to the lookout tower. It takes her 30 minutes to hike back down to the parking lot. Her speed going downhill is 1.2 miles per hour faster than her speed going uphill. Find Suzy’s uphill and downhill speeds.

uphill 1.8 mph, downhill three mph

## Try It \(\PageIndex{12}\)

Llewyn takes 45 minutes to drive his boat upstream from the dock to his favorite fishing spot. It takes him 30 minutes to drive the boat back downstream to the dock. The boat’s speed going downstream is four miles per hour faster than its speed going upstream. Find the boat’s upstream and downstream speeds.

upstream 8 mph, downstream 12 mph

In the distance, rate, and time formula, time represents the actual amount of elapsed time (in hours, minutes, etc.). If a problem gives us starting and ending times as clock times, we must find the elapsed time in order to use the formula.

## Example \(\PageIndex{13}\)

Hamilton loves to travel to Las Vegas, 255 miles from his home in Orange County. On his last trip, he left his house at 2:00 pm. The first part of his trip was on congested city freeways. At 4:00 pm, the traffic cleared and he was able to drive through the desert at a speed 1.75 times as fast as when he drove in the congested area. He arrived in Las Vegas at 6:30 pm. How fast was he driving during each part of his trip?

A diagram will help us model this trip.

Next, we create a table to organize the information.

We know the total distance is 255 miles. We are looking for the rate of speed for each part of the trip. The rate in the desert is 1.75 times the rate in the city. If we let r= the rate in the city, then the rate in the desert is 1.75r.

The times here are given as clock times. Hamilton started from home at 2:00 pm and entered the desert at 4:30 pm. So he spent two hours driving the congested freeways in the city. Then he drove faster from 4:00 pm until 6:30 pm in the desert. So he drove 2.5 hours in the desert.

Now, we multiply the rates by the times.

By looking at the diagram below, we can see that the sum of the distance driven in the city and the distance driven in the desert is 255 miles.

## Try It \(\PageIndex{14}\)

Cruz is training to compete in a triathlon. He left his house at 6:00 and ran until 7:30. Then he rode his bike until 9:45. He covered a total distance of 51 miles. His speed when biking was 1.6 times his speed when running. Find Cruz’s biking and running speeds.

biking 16 mph, running 10 mph

## Try It \(\PageIndex{15}\)

Phuong left home on his bicycle at 10:00. He rode on the flat street until 11:15, then rode uphill until 11:45. He rode a total of 31 miles. His speed riding uphill was 0.6 times his speed on the flat street. Find his speed biking uphill and on the flat street.

uphill 12 mph, flat street 20 mph

## Key Concepts

- D = rt where D = distance, r = rate, t = time
- Read the problem. Make sure all the words and ideas are understood. Draw a diagram to illustrate what it happening. Create a table to organize the information: Label the columns rate, time, distance. List the two scenarios. Write in the information you know.
- Name what we are looking for. Choose a variable to represent that quantity. Complete the chart. Use variable expressions to represent that quantity in each row. Multiply the rate times the time to get the distance.
- Translate into an equation. Restate the problem in one sentence with all the important information. Then, translate the sentence into an equation.

## Physics Problems with Solutions

Velocity and speed: solutions to problems.

Solutions to the problems on velocity and speed of moving objects. More tutorials can be found in this website.

A man walks 7 km in 2 hours and 2 km in 1 hour in the same direction. a) What is the man's average speed for the whole journey? b) What is the man's average velocity for the whole journey? Solution to Problem 1: a)

You start walking from a point on a circular field of radius 0.5 km and 1 hour later you are at the same point. a) What is your average speed for the whole journey? b) What is your average velocity for the whole journey? Solution to Problem 3: a) If you walk around a circular field and come back to the same point, you have covered a distance equal to the circumference of the circle.

John drove South 120 km at 60 km/h and then East 150 km at 50 km/h. Determine a) the average speed for the whole journey? b) the magnitude of the average velocity for the whole journey? Solution to Problem 4: a)

If I can walk at an average speed of 5 km/h, how many miles I can walk in two hours? Solution to Problem 5: distance = (average speed) * (time) = 5 km/h * 2 hours = 10 km using the rate of conversion 0.62 miles per km, the distance in miles is given by distance = 10 km * 0.62 miles/km = 6.2 miles

A train travels along a straight line at a constant speed of 60 mi/h for a distance d and then another distance equal to 2d in the same direction at a constant speed of 80 mi/h. a)What is the average speed of the train for the whole journey?

A car travels 22 km south, 12 km west, and 14 km north in half an hour. a) What is the average speed of the car? b) What is the final displacement of the car? c) What is the average velocity of the car?

Solution to Problem 7: a)

Solution to Problem 8: a)

## More References and links

- Velocity and Speed: Tutorials with Examples
- Velocity and Speed: Problems with Solutions
- Acceleration: Tutorials with Examples
- Uniform Acceleration Motion: Problems with Solutions
- Uniform Acceleration Motion: Equations with Explanations

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## Wind and Current Word Problems

Related Topics: More Lessons for Algebra II Math Worksheets

Videos, worksheets, solutions and activities to help Algebra 1 students learn how to solve wind and current word problems.

Rate of current problem #2 Shows how to solve a word problem involving the rate of a current and rowing in still water using 2 variables and 2 linear equations. Example: A dolphin swimming against an ocean current traveled 60 miles in 2 hours. It then turned around to swim with the current and was able to get back to its starting place in 1.5 hours. Find the speed of the dolphin in still water and the rate of the current.

Wind and Current Problems How to solve wind and current word problems using 2 variables and a system of linear equations? What happens to the rate when the wind is at your back? What happens when you try to paddle a kayak upstream? Examples: (1) A plane can fly 3750 km in 3 hours with the wind. The plane takes 5 hours to travel the same distance against the same wind speed. Find the rate of the plane in still air. Find the speed of the wind. (2) Jim can ow a boat 30 km downstream in 3 hours, but it takes him 5 hours to return. What is his rate in still water? What is the rate of the current?

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## Kinematics Practice Problems with Answers

Are you struggling with kinematics problems? Do you want to understand the principles of motion in a clear, concise manner? Look no further! Our comprehensive guide on “Kinematics Problems” is here to help.

Whether you’re studying for an exam or working on homework, these solutions offer a practical approach to understanding and applying kinematics equations.

All kinematics equations are summarized in the following expressions: \begin{gather*} \Delta x=\frac{v_1+v_2}{2}\times \Delta t \\\\ v=v_0+at \\\\ \Delta x=\frac 12 at^2+v_0t \\\\ v^2-v_0^2=2a\Delta x \end{gather*} In the rest of this long article, you will see how to apply these equations in the given problems.

If this article has helped you, consider donating to support our ongoing efforts to provide valuable resources for students like you.

## Kinematics Practice Problems:

Problem (1): A car slows down its motion from 10 m/s to 6 m/s in 2 seconds under constant acceleration. (a) What is its acceleration? (b) How far did the car travel during this time interval?

Solution : This is a basic kinematics problem, so we will explain the steps in detail.

Step 1: Since all these problems are in one dimension, draw a horizontal axis (like the positive $x$ axis), and place the object on it, so that its motion matches the direction of the axis.

Step 2: Specify the known and wanted information. Here, in the elapsed time interval $2\,{\rm s}$, the initial and final velocities of the car are given as $v_i=10\,{\rm m/s}$ and $v_f=6\,{\rm m/s}$. The wanted quantity is the constant acceleration of the object (car), $a=?$.

Step 3: Apply the kinematics equation that is appropriate for this situation.

(a) To find the acceleration in this problem, we are given the time, initial, and final velocity. The kinematics equation $v=v_0+at$ is suitable for this situation, as the only unknown variable is the acceleration $a$. By rearranging the equation, we get \begin{gather*} v=v_0+at\\\\ 6=10+a(2) \\\\ 6-10=2a \\\\\Rightarrow \quad a=\frac{6-10}{2}=-2\quad {\rm \frac{m}{s^2}}\end{gather*} Since the problem states that the acceleration is constant, we could also use any of the other constant acceleration kinematics equations. The negative sign of the acceleration indicates that it is directed toward the negative $x$ axis.

(b) "How far'' indeed refers to the distance traveled, denoted by $x$ in the kinematics equations.

Here, the best equation that relates the known and unknown information is $x=\frac 12 at^2+v_0t$ or $v^2-v_0^2=2ax$. We choose the first, so \begin{align*} x&=\frac 12 at^2+v_0t \\\\&=\frac 12 (-2)(2)^2+(10)(2) \\\\&=16\quad {\rm m}\end{align*}

On the following page you can find over 40+ questions related to applying kinematics equations in velocity and acceleration:

Velocity and acceleration problems

Problem (2): A moving object slows down from $12\,{\rm m/s}$ to rest at a distance of 20 m. Find the acceleration of the object (assumed constant).

Solution : In the diagram below, all known information along with the direction of the uniform motion is shown.

A common phrase in kinematics problems is “ending or coming to a rest”, which means the final velocity of the object in the time interval we are considering is zero, $v_f=0$.

The kinematics equation that suits this problem is $v^2-v_0^2=2a(x-x_0)$, where the only unknown variable is the acceleration $a$.

For simplicity, we can assume the initial position of the motion $x_0$ is zero in all kinematics problems, $x_0$. \begin{gather*} v^2-v_0^2=2ax\\\\0^2-(12)^2 =2a(20) \\\\ \rightarrow a=\frac{-144}{2\times 20}\\\\\Rightarrow \boxed{a=-3.6\quad {\rm \frac{m}{s^2}}}\end{gather*} As before, the negative sign indicates of the acceleration indicates that it is directed to the left .

Problem (3): A bullet leaves the muzzle of an 84-cm rifle with a speed of 521 m/s. Find the magnitude of the bullet's acceleration by assuming it is constant inside the barrel of the rifle.

Solution : The bullet accelerates from rest to a speed of 521 m/s over a distance of 0.84 meters. We have the following known quantities: the initial velocity $v_0$, the final velocity $v$, and the displacement $x-x_0$. The unknown is acceleration $a$. The perfect kinematics equation that relates all these variables is $v^2-v_0^2=2a(x-x_0)$. Solving for $a$, we have: \begin{gather*}v^2-v_0^2=2a(x-x_0)\\\\ (521)^2-0=2(a)(0.84-0) \\\\ \Rightarrow \boxed{a=1.62\times 10^5\quad {\rm m/s^2}}\end{gather*} This calculation results in a very large acceleration.

Problem (4): A car starts its motion from rest and uniformly accelerates at a rate of $4\,{\rm m/s^2}$ for 2 seconds in a straight line. (a) How far did the car travel during those 2 seconds? (b) What is the car's velocity at the end of that time interval?

Solution : "Start from rest'' means the initial object's velocity is zero, $v_0=0$. The known information are $a=4\,{\rm m/s^2}$, $t=2\,{\rm s}$ and wants the distance traveled $x=?$.

(a) The kinematics equation that relates the given information is $x=\frac 12 at^2+v_0 t+x_0$ since the only unknown quantity is $x$. Given the known data, we can calculate $x$ as follows: \begin{align*} x&=\frac 12 at^2+v_0 t+x_0 \\\\&=\frac 12 (4)(2)^2+(0)(2) \\\\&=\boxed{8\quad {\rm m}}\end{align*} As before, we set $x_0=0$.

(b) Now that we know the distance traveled by car in that time interval, we can use the following kinematics equation to find the car's final velocity $v$: \begin{align*} v^2-v_0^2 &=2a(x-x_0) \\\\v^2-(0)^2&=2(4)(8-0) \\\\v^2&=64\end{align*} Taking the square root, we get $v$: \[v=\sqrt{64}=\pm 8\quad {\rm \frac ms}\] We know that velocity is a vector quantity in physics and has both a direction and a magnitude.

The magnitude of the velocity (speed) was obtained as 8 m/s, but in what direction? Or which sign should we choose? Because the car is uniformly accelerating without stopping in the positive $x$ axis, the correct sign for velocity is positive.

Therefore, the car's final velocity is $\boxed{v_f=+8\,{\rm m/s}}$.

Problem (5): We aim to design an airport runway with the following specifications: The lowest acceleration of a plane should be $4\,{\rm m/s^2}$, and its take-off speed should be 75 m/s. How long would the runway have to be to allow the planes to accelerate through it?

Solution: The known quantities are acceleration $a=4\,{\rm m/s^2}$, and final velocity $v=75\,{\rm m/s}$. The wanted quantity is the runway length $\Delta x=x-x_0$. The ideal kinematics equation that relates these variables is $v^2-v_0^2=2a(x-x_0)$. \begin{align*} v^2-v_0^2&=2a\Delta x\\\\ (75)^2-0&=2(4) \Delta x\\\\ \Rightarrow \Delta x&=\boxed{703\quad {\rm m}}\end{align*} Thus, for the runway to be effective, its length must be at least approximately 703 meters.

Problem (6): A stone is dropped vertically from a high cliff. After 3.55 seconds, it hits the ground. How high is the cliff?

Solution : There is another type of kinematics problem in one dimension but in the vertical direction. In such problems, the constant acceleration is that of free falling, $a=g=-10\,{\rm m/s^2}$.

"Dropped'' or "released'' in free-falling problems means the initial velocity is zero, $v_0=0$. In addition, it is always better to consider the point of release as the origin of the coordinate, so $y_0=0$.

The most relevant kinematics equation for these known and wanted quantities is $y=-\frac 12 gt^2+v_0t+y_0$ \begin{align*} y&=-\frac 12 gt^2+v_0t+y_0 \\\\&=-\frac 12 (9.8)(3.55)^2+0+0\\\\&=\boxed{-61.8\quad {\rm m}}\end{align*} The negative indicates that the impact point is below our chosen origin .

Problem (7): A car slows down uniformly from $45\,\rm m/s$ to rest in $10\,\rm s$. How far did it travel in this time interval?

Solution : List the data known as follows: initial speed $v_0=45\,\rm m/s$, final speed $v=0$, and the total time duration that this happened is $t=10\,\rm s$. The unknown is also the amount of displacement, $\Delta x$.

The only kinematics equation that relates these together is $\Delta x=\frac{v_1+v_2}{2}\times \Delta t$, where $v_1$ and $v_2$ are the velocities at the beginning and end of that time interval. \begin{align*} \Delta x&=\frac{v_1+v_2}{2}\times \Delta t \\\\ &=\frac{45+0}{2}\times 10 \\\\ &=\boxed{225\,\rm m}\end{align*} Keep in mind that we use this formula when the object slows down uniformly, or, in other words when the object's acceleration is constant.

Problem (8): A ball is thrown into the air vertically from the ground level with an initial speed of 20 m/s. (a) How long is the ball in the air? (b) At what height does the ball reach?

Solution : The throwing point is considered to be the origin of our coordinate system, so $y_0=0$. Given the initial velocity $v_0=+20\,{\rm m/s}$ and the gravitational acceleration $a=g=-9.8\,{\rm m/s^2}$. The wanted time is how long it takes the ball to reach the ground again.

To solve this free-fall problem , it is necessary to know some notes about free-falling objects.

Note (1): Because the air resistance is neglected, the time the ball is going up is half the time it is going down.

Note (2): At the highest point of the path, the velocity of the object is zero.

(a) By applying the kinematics equation $v=v_0+at$ between the initial and the highest ($v=0$) points of the vertical path, we can find the going up time. \begin{align*} v&=v_0+at \\0&=20+(-9.8)t\\\Rightarrow t&=2.04\quad {\rm s}\end{align*} The total flight time is twice this time \[t_{tot}=2t=2(2.04)=4.1\,{\rm s}\] Hence, the ball takes about 4 seconds to reach the ground.

(b) The kinematics equation $v^2-v_0^2=2a(y-y_0)$ is best for this part. \begin{align*} v^2-v_0^2&=2a(y-y_0) \\0-20^2&=2(-9.8)(y-0) \\ \Rightarrow y&=\boxed{20\quad {\rm m}}\end{align*} Hence, the ball goes up to a height of about 20 meters.

Problem (9): An object moving in a straight line with constant acceleration, has a velocity of $v=+10\,{\rm m/s}$ when it is at position $x=+6\,{\rm m}$ and of $v=+15\,{\rm m/s}$ when it is at $x=10\,{\rm m}$. Find the acceleration of the object.

Solution : Draw a diagram, put all known data into it, and find a relevant kinematics equation that relates them together.

We want to analyze the motion in a distance interval of $\Delta x=x_2-x_1=10-6=4\,{\rm m}$, thus, we can consider the velocity at position $x_1=6\,{\rm m}$ as the initial velocity and at $x_2=10\,{\rm m}$ as the final velocity.

The most relevant kinematics equation that relates these known quantities to the wanted acceleration $a$ is $v^2-v_0^2=2a(x-x_0)$, where $x-x_0$ is the same given distance interval. Thus, \begin{align*} v^2-v_0^2&=2a(x-x_0) \\\\ (15)^2-(10)^2&=2(a)(4) \\\\225-100&=8a\\\\\Rightarrow a&=\frac{125}{8}\\\\&=15.6\,{\rm m/s^2}\end{align*}

Problem (10): A moving object accelerates uniformly from 75 m/s at time $t=0$ to 135 m/s at $t=10\,{\rm s}$. How far did it move at the time interval $t=2\,{\rm s}$ to $t=4\,{\rm s}$?

Solution : Draw a diagram and implement all known data in it as below.

Because the problem tells us that the object accelerates uniformly, we can infer that its acceleration is constant throughout its entire path.

Given the initial and final velocities of the moving object, we can determine its acceleration using the definition of instantaneous acceleration as follows: \[a=\frac{v_2-v_1}{t_2-t_1}=\frac{135-75}{10}=6\,{\rm m/s^2}\] To analyze the motion between the requested times (referred to as stage II in the figure), we need some information for that time interval, such as their velocities or the distance between them.

As you can see in the figure, the initial velocity of stage II is the final velocity of stage I. By using a relevant kinematics equation that relates these data to each other, we find that \begin{align*} v&=v_0+at\\\\&=75+(6)(2) \\\\&=87\,{\rm m/s}\end{align*} This velocity will be the initial velocity for stage II of the motion. Now, all the known information for stage II is as follows: initial velocity $v_0=87\,{\rm m/s}$, acceleration $a=6\,{\rm m/s^2}$, and time interval $\Delta t=2\,{\rm s}$. The unknown is the distance traveled, denoted by $x=?$

The appropriate equation that relates all these variables is $x=\frac 12 at^2+v_0t+x_0$. Substituting the known values, we get \begin{align*}x&=\frac 12 at^2+v_0t+x_0\\\\&=\frac 12 (6)(2)^2+(87)(2)+0\\\\&=186\quad {\rm m}\end{align*} Hence, our moving object travels a distance of 186 m between the instances of 2 s and 4 s.

Problem (11): A fast car starts from rest and accelerates at a uniform rate of $1.5\,{\rm m/s^2}$ for 4 seconds. After a while, the driver applies the brakes for 3 seconds, causing the car to uniformly slow down (decelerate) at a rate of $-2\,{\rm m/s^2}$. (a) How fast is the car at the end of the braking period? (b) How far has the car traveled after the braking period?

Solution : This motion is divided into two parts. First, draw a diagram and specify each section's known kinematics quantities.

(a) In the first part, given the acceleration, initial velocity, and time interval, we can calculate the final velocity at the end of 4 seconds. \begin{align*} v&=v_0+at\\&=0+(1.5)(4) \\&=6\quad {\rm m/s}\end{align*} This velocity is considered as the initial velocity for the second part, where we want to find the final velocity.

In the next part, given the magnitude of acceleration and braking time interval, we can calculate the final velocity as follows: \begin{align*} v&=v_0+at\\&=6+(-2)(3) \\&=0\end{align*} The zero velocity here indicates that the car comes to a stop after the braking period.

(b) The distance traveled in the second part can now be calculated using the kinematics equation $x=\frac 12 at^2+v_0t+x_0$, because the only unknown quantity is distance $x$. \begin{align*} x&=\frac 12 at^2+v_0t+x_0\\\\&=\frac 12 (-2)(3)^2+(6)(3)+0\\\\&=+9\quad {\rm m}\end{align*} Therefore, after braking, the car travels a distance of 9 meters before coming to a stopped.

Problem (12): A car moves at a speed of 20 m/s down a straight path. Suddenly, the driver sees an obstacle in front of him and applies the brakes. Before the car reaches a stop, it experiences an acceleration of $-10\,{\rm m/s^2}$. (a) After applying the brakes, how far did it travel before stopping? (b) How long does it take the car to reach a stop?

Solution : As always, the first and most important step in solving a kinematics problem is to draw a diagram and input all known values into it, as shown below.

(a) The kinematics equation $v^2-v_0^2=2a(x-x_0)$ is the perfect equation to use here, as the only unknown quantity in it is the distance traveled, denoted by $x$. Thus, \begin{align*} v^2-v_0^2&=2a(x-x_0) \\\\ 0^2-(20)^2&=2(-10)(x-0) \\\\\Rightarrow \quad x&=\frac {-400}{-20}\\\\&=20\quad {\rm m}\end{align*} (b) The phrase "How long does it take'' asks us to find the time interval. The initial and final velocities, as well as acceleration, are known, so the only relevant kinematics equation is $v=v_0+at$. Thus, \begin{align*} v&=v_0+at\\\\0&=20+(-10)t\\\\\Rightarrow t&=\frac{-20}{-10}\\\\&=2\quad {\rm s}\end{align*} Therefore, after braking, the car moves for 2 seconds before coming to a stop.

Problem (13): A sports car moves a distance of 100 m in 5 seconds with a uniform speed. Then, the driver brakes, and the car, comes to a stop after 4 seconds. Find the magnitude and direction of its acceleration (assumed constant).

Solution : uniform speed means constant speed or zero acceleration for the motion before braking. Thus, we can use the definition of average velocity to find its speed just before braking as below \begin{align*} \bar{v}&=\frac{\Delta x}{\Delta t}\\\\&=\frac{100}{5}\\\\&=20\quad {\rm m/s}\end{align*} Now, we know the initial and final velocities of the car in the braking stage. Since the acceleration is assumed to be constant, by applying the definition of average acceleration, we would have \begin{align*} \bar{a}&=\frac{v_2-v_1}{\Delta t}\\\\&=\frac{0-20}{4}\\\\&=-5\quad {\rm m/s^2}\end{align*} The negative shows the direction of the acceleration, which is toward the negative $x$-axis.

Hence, the car's acceleration has a magnitude of $5\,{\rm m/s^2}$ in the negative $x$ direction.

Problem (14): A race car accelerates from rest at a constant rate of $2\,{\rm m/s^2}$ in 15 seconds. It then travels at a constant speed for 20 seconds, and after that, it comes to a stop with an acceleration of $2\,{\rm m/s^2}$. (a) What is the total distance traveled by car? (b) What is its average velocity over the entire path?

Solution : To solve this kinematics question, we divided the entire path into three parts.

Part I: "From rest'' means the initial velocity is zero. Thus, given the acceleration and time interval, we can use the kinematics equation $v=v_0+at$ to calculate the distance traveled by the car at the end of 15 seconds for the first part of the path. \begin{align*} x&=\frac{1}{2}at^2 +v_0 t+x_0\\\\&=\frac 12 (2)(15)^2 +(0)(15)+0\\\\&=\boxed{125\quad{\rm m}}\end{align*} As a side calculation, we find the final velocity for this part as below \begin{align*}v&=v_0+at\\\\&=0+(2) (15) \\\\&=30\quad {\rm m/s}\end{align*} Part II: the speed in this part is the final speed in the first part because the car continues moving at this constant speed after that moment.

The constant speed means we are facing zero acceleration. As a result, it is preferable to use the average velocity definition rather than the kinematics equations for constant (uniform) acceleration.

The distance traveled for this part, which takes 20 seconds at a constant speed of 30 m/s, is computed by the definition of average velocity as below \begin{align*} \bar{v}&=\frac{\Delta x}{\Delta t}\\\\30&=\frac{\Delta x}{20}\end{align*} Thus, we find the distance traveled as $\boxed{x=600\,{\rm m}}$.

Part III: In this part, the car comes to a stop, $v=0$, so its acceleration must be a negative value as $a=-2\,{\rm m/s^2}$. Here, the final velocity is also zero. Its initial velocity is the same as in the previous part.

Consequently, the best kinematics equation that relates those known to the wanted distance traveled $x$, is $v^2-v_0^2=2a(x-x_0)$. \begin{align*} v^2-v_0^2&=2a(x-x_0) \\\\0^2-(30)^2&=2(-2)(x-0) \\\\ \Rightarrow \quad x&=\boxed{225\quad {\rm m}}\end{align*} The total distance traveled by car for the entire path is the sum of the above distances \[D=125+600+225=\boxed{950\quad {\rm m}}\]

Problem (15): A ball is dropped vertically downward from a tall building of 30-m-height with an initial speed of 8 m/s. After what time interval does the ball strike the ground? (take $g=-10\,{\rm m/s^2}$.)

Solution : This is a free-falling kinematics problem. As always, choose a coordinate system along with the motion and the origin as the starting point.

Here, the dropping point is considered the origin, so in all kinematics equations, we set $y_0=0$. By this choice, the striking point is 30 meters below the origin, so in equations, we also set $y=-30\,{\rm m}$.

Remember that velocity is a vector in physics whose magnitude is called speed. In this problem, the initial speed is 8 m/s downward. This means that the velocity vector is written as $v=-8\,{\rm m/s}$.

Now that all necessary quantities are ready, we can use the kinematics equation $y=\frac 12 at^2+v_0t+y_0$, to find the wanted time that the ball strikes the ground. \begin{align*} y&=\frac 12 at^2+v_0t+y_0\\\\-30&=\frac 12 (-10)t^2+(-8)t+0\end{align*} After rearranging, a quadratic equation like $5t^2+8t-30=0$ is obtained, whose solutions are given as below: \begin{gather*} t=\frac{-8\pm\sqrt{8^2-4(5)(-30)}}{2(5)}\\\\ \boxed{t_1=1.77\,{\rm s}} \quad , \quad t_2=-3.37\,{\rm s}\end{gather*} $t_1$ is the accepted time because the other is negative, which is not acceptable in kinematics. Therefore, the ball takes about 1.7 seconds to hit the ground.

Note: The solutions of a quadratic equation like $at^2+bt+c=0$, where $a,b,c$ are some constants, are found by the following formula: \[t=\frac{-b\pm\sqrt{b^2-4ac}}{2a}\]

Problem (16): The acceleration versus time graph for an object that moves at a constant speed of 30 m/s is shown in the figure below. Find the object's average velocity between instances $t_1=10\,{\rm s}$ and $t_2=30\,{\rm s}$.

Solution : The best and shortest approach to solving such a kinematics problem is to first draw its velocity-vs-time graph. Next, the area under the obtained graph gives us the total displacement, which is divided by the total time interval to yield the average velocity.

The path consists of three parts with different accelerations.

In the first part, the object slows down its motion at a constant rate of $-2\,{\rm m/s^2}$ in 10 seconds. Its initial velocity is also 30 m/s. With these known quantities in hand, the kinematics equation $v=v_0+at$ gives us the velocity at the end of this time interval. \begin{align*} v&=v_0+at\\&=30+(-2)(10) \\&=10\quad {\rm m/s}\end{align*} This calculation corresponds to a straight line between the points $(v=30\,{\rm m/s},t=0)$ and $(v=10\,{\rm m/s},t=10\,{\rm s})$ on the $v-t$ graph as shown below.

Next, the object moves with zero acceleration for 5 seconds, which means the velocity does not change during this time interval. This implies that we must draw a horizontal line in the $v-t$ graph.

In the last part, the object accelerates from 10 m/s with a constant rate of $+2\,{\rm m/s^2}$ in 15 seconds. Thus, its final velocity at the end of this time interval is determined as below \begin{align*} v&=v_0+at\\&=10+(2)(15) \\&=40\quad {\rm m/s}\end{align*} Now, it's time to draw the velocity-vs-time graph. As an important point, note that all these motions have a constant acceleration, so all parts of a velocity-time graph, are composed of straight-line segments with different slopes.

For part I, we must draw a straight-line segment between the velocities of 30 m/s and 10 m/s.

Part II is a horizontal line since its velocities are constant during that time interval, and finally, in Part III, there is a straight line between velocities of 10 m/s and 40 m/s.

All these verbal phrases are illustrated in the following velocity-vs-time graph .

Recall that the area under a velocity vs. time graph always gives the displacement. Hence, the area under the $v-t$ graph between 10 s and 30 s gives the displacement. Therefore, the areas of rectangle $S_1$ and trapezoid $S_2$ are calculated as below \begin{gather*} S_1 =10\times 5=50\quad {\rm m} \\\\S_2=\frac{10+40}{2}\times 15=375\quad {\rm m}\end{gather*} Therefore, the total displacement in the time interval $[15,30]$ is \[D=S_{tot}=S_1+S_2=425\,{\rm m}\] From the definition of average velocity, we have \[\bar{v}=\frac{displacement}{time}=\frac{425}{20}=21.25\,{\rm m/s}\]

## Challenging Kinematics Problems

In the following, some challenging kinematics problems are presented for homework.

A driver is moving along at $45\,\rm m/s$ when she suddenly notices a roadblock $100\,\rm m$ ahead. Can the driver stop the vehicle in time to avoid colliding with the obstruction if her reaction time is assumed to be $0.5\,\rm s$ and her car's maximum deceleration is $5\,\rm m/s^2$?

Solution : The time between seeing the obstacle and taking action, such as slamming on the brake, is defined as the reaction time. During this time interval, the moving object travels at a constant speed.

Thus, in all such questions, we have two phases. One is constant speed, and the other is accelerating with negative acceleration (deceleration).

Here, between the time of seeing the barrier and the time of braking, the driver covers a distance of \begin{align*} x_1&=vt_{reac} \\\\ &=25\times 0.5 \\\\ &=12.5\,\rm m\end{align*} In the decelerating phase, the car moves a distance, which is obtained using the following kinematics equation: \begin{gather*} v^2-v_0^2=2ax_2 \\\\ (0)^2-(25)^2=2(-5) x_2 \\\\ \Rightarrow x_2=62.5\,\rm m\end{gather*} Summing these two distances gives a total distance that is a good indication of whether the moving object hits the obstacle or not. \begin{gather*} \Delta x_{actual}=x_1+x_2=75\,\rm m \\\\ \Rightarrow \Delta x_{covered}<\Delta x_{actual} \end{gather*} As a result, because the distance covered by the car is less than the actual distance between the time of seeing the barrier and the obstacle itself, the driver has sufficient time to stop the car in time to avoid a collision.

For a moving car at a constant speed of $90\,\rm km/h$ and a human reaction time of $0.3\,\rm s$; find the stopping distance if it slows down at a rate of $a=3\,\rm m/s^2$.

Solution : We use SI units, so first convert the given speed in these units as below \begin{align*} v&=90\,\rm km/h \\\\ &=\rm 90\times \left(\frac{1000\,m}{3600\,s}\right) \\\\ &=25\,\rm m/s\end{align*} As we said previously, during the reaction time, your car moves at a constant speed and covers a distance of \begin{align*} x_1&=vt_{react} \\\\ &=(25)(0.3) \\\\ &=7.5\,\rm m \end{align*} Deceleration means the moving object slows down, or a decrease per second in the velocity of the car occurs. In this case, we must put the acceleration with a negative sign in the kinematics equations.

During the second phase, your car has negative acceleration and wants to be stopped. Thus, the distance covered during this time interval is found as follows \begin{align*} v^2-v_0^2=2a\Delta x \\\\ (0)^2-(25)^2=2(3)\Delta x \\\\ \Rightarrow \quad \boxed{\Delta x=104.17\,\rm m}\end{align*}

Assume you are traveling $35\,\rm m/s$ when suddenly you see red light traffic $50\,\rm m$ ahead. If it takes you $0.456\,\rm s$ to apply the brakes and the maximum deceleration of the car is $4.5\,\rm m/s^2$, (a) Will you be able to stop the car in time? (b) How far from the time of seeing the red light will you be?

Solution : When you see the red light until you apply the brakes, your car is moving at a constant speed. This time interval is defined as the reaction time, $\Delta t_{react}=0.456\,\rm s$. After you get the brakes on, the car starts to decelerate at a constant rate, $a=-4.5\,\rm m/s^2$. Pay attention to the negative signs of such problems. The negative tells us that the car is decreasing its speed.

(a) In the first phase, the car moves a distance of \begin{align*} x_1&=v\Delta t_{react} \\\\ &=35\times 0.455 \\\\ &=15.96\,\rm m\end{align*} In the phase of deceleration, the car is moving a distance whose magnitude is found using the time-independent kinematics equation as below \begin{gather*} v^2-v_0^2=2ax_2 \\\\ (0)^2-(35)^2=2(4.5)x_2 \\\\ \Rightarrow \quad x_2=136.11\,\rm m\end{gather*} The sum of these two distances traveled gives us the total distance covered by the car from the time of seeing the red traffic light to the moment of a complete stop. \[x_{tot}=x_1+x_2=152.07\,\rm m \] Because the total distance traveled is greater than the actual distance to the red light, the driver will not be able to stop the car in time.

(b) As previously calculated, the total distance traveled by the car is nearly $152\,\rm m$ or the car is about $102\,\rm m$ past the red light traffic.

A person stands on the edge of a $60-\,\rm m$-high cliff and throws two stones vertically downward, $1$ second apart, and sees they both reach the water simultaneously. The first stone had an initial speed of $4\,\rm m/s$. (a) How long after dropping the first stone does the second stone hit the water? (b) How fast was the second stone released? (c) What is the speed of each stone at the instant of hitting the water?

Solution: Because all quantities appearing in the kinematics equation are vectors, we must first choose a positive direction. Here, we take up as a positive $y$ direction.

Both stones arrived in the water at the same time. Thus, calculate the time the first stone was in the air. Next, use the time interval between the two drops to find the duration the second stone was in the air. (a) The first stone is released downward at a speed of $4\,\rm m/s$, thus, its initial velocity is $v_0=-4\,\rm m/s$. The minus sign is for moving in the opposite direction of the chosen direction.

The only relevant kinematics equation that relates this known information is $\Delta y=-\frac 12 gt^2+v_0t$, where $\Delta y=-60\,\rm m$ is the vertical displacement, and the negative indicates that the stone hit a point below the chosen origin. Substituting the numerical values into this and solving for the time duration $t$ gives \begin{gather*} \Delta y=-\frac 12 gt^2+v_0t \\\\ -60=-\frac 12 (10)t^2+(-4)t \\\\ 5t^2+4t-60=0 \\\\ \Rightarrow \boxed{t=3.0\,\rm s} \, , \, t'=-3.8\,\rm s \end{gather*} The second answer is not acceptable.

(b) The second stone was released $1$ second after throwing the first one and arrived at the same time as the first stone. Therefore, the time interval that the second stone was in the air is found to be \begin{align*} t_2&=t_1-1 \\ &=3.0-1\\ &=2\,\rm s\end{align*}

(c) It is better to apply the time-independent kinematics equation $v^2-v_0^2=-2g\Delta y$ to find the stone's velocity at the moment it hit the water. For the first stone, we have \begin{gather*} v^2-v_0^2=-2g\Delta y \\\\ v^2-(-4)^2=-2(10)(-60) \\\\ \Rightarrow \quad \boxed{v=34.8\,\rm m/s} \end{gather*} The second stone's velocity is left to you as an exercise.

In a tennis game, the ball leaves the racket at a speed of $75\,\rm m/s$ whereas it is in contact with the racket for $25\,\rm ms$, and starts at rest. Assume the ball experiences constant acceleration. (a) What was the ball's acceleration during this serve? (b) How far has the ball traveled on this serve?

Solution : In this question, we are asked to find the ball's acceleration and distance traveled during that pretty small time interval. (a) In a time interval of $\Delta t=25\times 10^{-3}\,\rm s$, we are given the beginning velocity $v_1=0$ and the end velocity $v_2=85\,\rm m/s$. Because it is assumed the acceleration is constant, the average acceleration definition, $a=\frac{\Delta v}{\Delta t}$, is best suited for these known quantities. \begin{align*} a&=\frac{v_2-v_1}{\Delta t} \\\\ &=\frac{75-0}{25\times 10^{-3}} \\\\ &=3000\,\rm m/s^2 \end{align*} A huge acceleration is given to the tennis ball. (b) Here, we are asked to find the amount of distance traveled by the ball during the time the ball was in contact with the racket. Because we have a constant acceleration motion, it is best to use the following equation to find the distance traveled. \begin{align*} \Delta x&=\frac{v_1+v_2}{2}\times \Delta t \\\\ &=\frac{0+75}{2}\times (25\times 10^{-3}) \\\\ &=937.5\times 10^{-3}\,\rm m\end{align*} In millimeters, $\Delta x=937.5\,\rm mm$, and in centimeters $\Delta x=93.75\,\rm cm$. Therefore, during this incredibly short time interval, the ball moves about $94\,\rm cm$ along with the racket.

Starting from rest and ending at rest, a car travels a distance of $1500\,\rm m$ along the $x$-axis. During the first quarter of the distance, it accelerates at a rate of $+1.75\,\rm m/s^2$, while for the remaining distance, its acceleration is $-0.450\,\rm m/s^2$. (a) What is the time travel of the whole path? (b) What is the maximum speed of the car over this distance?

None of the time-dependent kinematics equations give us the time travel $t'$ without knowing the initial speed at the instant of the start of this second path.

We can find it using the equation $v^2-v_0^2=2a\Delta x$, setting $v=0$ at the end of the path, and solving for $v_0$ \begin{gather*} v^2-v_0^2=2a\Delta x \\\\ (0)^2-v_0^2=2(-0.450)(1125) \\\\ v_0=\sqrt{2\times 0.45\times 1125} \\\\ \Rightarrow v_0=31.82\,\rm m/s\end{gather*} Given that, one can use the simple equation $v=v_0+at$ and solve for the time travel in this part of the path. \begin{gather*} v=v_0+at \\\\ 0=31.82+(-0.450)t' \\\\ \Rightarrow t'=70.71\,\rm s\end{gather*} Therefore, the total time traveled over the entire path is the sum of these two times. \begin{align*} T&=t+t' \\\\ &=20.70+70.71 \\\\ &=\boxed{91.41\,\rm s} \end{align*}

A train that is $75$ meters long starts accelerating uniformly from rest. When the front of the train reaches a railway worker who is standing $150$ meters away from where the train started, it is traveling at a speed of $20\,\rm m/s$. What will be the speed of the last car as it passes the worker?

Solution : The front of the train is initially $150\,\rm m$ away from the worker, and when it passes him, it has a speed of $25\,\rm m/s$. From this data, we can find the acceleration of the front of the train (which is the same acceleration as the whole train) by applying the following kinematics equation \begin{gather*} v^2-v_0^2=2a\Delta x \\\\ (25)^2-(0)^2=2a\times 150 \\\\ \Rightarrow \quad a=2.08\,\rm m/s^2\end{gather*} Given the train's acceleration, now focus on the last car.

The last car is initially at rest and placed at a distance of $150+80=230\,\rm m$ away from the person. When it passes the person, it has traveled $\Delta x= 230\,\rm m$ and its speed is determined simply as below \begin{gather*} v^2-v_0^2=2a\Delta x \\\\ v^2-(0)^2=2(2.08)(230) \\\\ \Rightarrow \quad \boxed{v=30.93\,\rm m/s}\end{gather*}

A wildcat moving with constant acceleration covers a distance of $100\,\rm m$ apart in $8\,\rm s$. Assuming that its speed at the second point is $20\,\rm m/s$, (a) What was its speed in the first place? (b) At what rate does its speed change over this distance?

Solution : First of all, list all known data given to us. Time interval $\Delta t=8\,\rm s$, the horizontal displacement $\Delta x=100\,\rm m$, speed at second point $v_2=20\,\rm m/s$.

We are asked to find the speed at the second point. To solve this kinematics problem, we use the following kinematics equation because the acceleration is constant and this is the most relevant equation that relates the known to the unknown quantities. \begin{gather*} \Delta x=\frac{v_1+v_2}{2}\times \Delta t \\\\ 100=\frac{v_1+20}{2}\times 8 \\\\ \Rightarrow \quad \boxed{v_1=5\,\rm m/s} \end{gather*} Therefore, the wildcat's speed in the first place is $5\,\rm m/s$.

In this part, we should find the wildcat's acceleration because acceleration is defined as the time rate of change of the speed of a moving object. Given the first place speed, $v_1=5\,\rm m/s$, found in the preceding part, we can use the following time-independent kinematics equation to find the wanted unknown. \begin{gather*} v_2^2-v_1^2=2a\Delta x \\\\ (20)^2-(5)^2=2a(100) \\\\ \Rightarrow \quad \boxed{a=1.875\,\rm m/s^2}\end{gather*}

In this tutorial, all concepts about kinematics equations are taught in a problem-solution strategy.

We can also find these kinematic variables using a position-time or velocity-time graph. Because slopes in those graphs represent velocity and acceleration, respectively, and the concavity of a curve in a position vs. time graph shows the sign of its acceleration in an x-t graph as well.

These multiple-choice questions on kinematics for AP Physics 1 are also available to review for students enrolled in AP Physics courses.

Author : Dr. Ali Nemati Date published : 8-7-2021 Updated : June 12, 2023

© 2015 All rights reserved. by Physexams.com

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## Waves, Sound and Light: Sound and Music

Calculator pad, version 2, sound and music: problem set.

The speed ( v ) at which sound travels through air is dependent upon the temperature of the air and seems to follow the equation v = 331 m/s + 0.6 m/s/°C * T where T is the Celsius temperature of the air. Determine the speed of sound … a. … on a cold day when the outdoor temperature is 4°C. b. … inside the school where the temperature is 24°C. c. … on a warm day when the outdoor temperature is 38°C.

- Audio Guided Solution
- Show Answer

a. 333 m/s b. 345 m/s c. 354 m/s

Herds of African elephants are generally spread over large areas. Infrasonic sound waves (sound waves below the human range of frequency detection) are used by these elephants to locate each other and to communicate. Sound waves with low frequencies have a greater ability to bend around obstacles and generally carry further. Scientists have detected sound waves with frequencies as low as 13 Hz being produced by elephants. Assuming a speed of sound of 350 m/s, determine the wavelength of these sounds waves.

Indoor pests such as mice and other rodents are sensitive to ultrasonic sound waves (sound waves above the human range of frequency detection). Some companies have produced (allegedly) rodent repellant devices that emit ultrasonic waves with frequencies of approximately 45 kHz. Assuming a speed of sound of 344 m/s, determine the wavelength of these sounds waves.

Mama G used to be the leader of South's Pep Club. As she started in on a round of Wash Them Down the River , her voice would ring out at an attention-drawing 855 Hz. Assuming a speed of sound in the Titan Dome of 355 m/s, determine the wavelength of the sound waves produced by Mama G.

On a recent PE-sponsored adventure education program, students went hiking at Devil's Head State Park. At one point, Jeremy let out a holler which reflected off a nearby rocky cliff and was detected as an echo 1.80 seconds later. Determine the distance to the rocky cliffs. Assume a speed of sound of 344 m/s.

A deep sea ocean vessel uses SONAR to detect the ocean's bottom. Sound waves are emitted from the surface of the ocean and travel through the water at 1450 m/s. The ocean bottom is 1630 m below the surface. Determine the amount of time that passes before the sound waves are reflected back to the surface.

The intensity of sound waves decreases as the distance from the source of sound increases. The relationship between intensity ( I ) and distance ( d ) is an inverse square relationship which follows the equation I = P/(4•π•R 2 ) where P is the power of the sound source, usually expressed in Watts. Jake recently purchased a stereo system for his basement recreation room. Determine the maximum intensity of the sound waves at the following distances from his 120-Watt main speaker. a. 1.0 meter b. 2.0 meter c. 3.0 meter

a. 9.5 W/m 2 b. 2.4 W/m 2 c. 1.1 W/m 2

Determine the decibel rating of the following sound sources and their estimated sound intensities. a. Science office at 5 PM on a weeknight: I = ~1 x 10 -9 W/m 2 b. South's student library after school: I = ~1 x 10 -6 W/m 2 c. Period 7 at the beginning of class: I = ~1 x 10 -4 W/m 2 d. Titan Dome on a Friday night during basketball season: I = 8.1 x 10 -3 W/m 2 e. Fall Out Boy concert - front row: I = 7.4 x 10 -2 W/m 2

a. 30 dB b. 60 dB c. 80 dB d. 99 dB e. 109 dB

For the following decibel levels, determine the corresponding sound intensity levels in W/m 2 . a. 50 dBel b. 90 dBel c. 110 dBel

a. 1.0 x 10 -7 W/m 2 b. 1.0 x 10 -3 W/m 2 c. 1.0 x 10 -1 W/m 2

## Problem 10:

According to Guinness, the record for the loudest burp is held by Paul Hunn of London. In September of 2008, his burp was measured at 107.1 dB, Determine the intensity in W/m 2 of Paul's burp.

## Problem 11:

Mr. H recently purchased a home four blocks from the busy tollway. On a typical evening, the decibel level resulting from tollway traffic is 62 dB at the location of his house. Determine the decibel level on the same evening at a house 1 block from the tollway (four times closer).

## Problem 12:

During a Variety Show practice, Jake plucked a string on his guitar, sending vibrations through it in both directions. The string is pulled to a tightness of 220 N and has a mass density of 0.013 kg/m. Determine the speed with which vibrations travel through the string.

## Problem 13:

During the Pluck It! Lab, lab partners Anna Litical and Noah Formula determined the speed of vibrations through a 2.45-meter length of wire. The wire had a mass of 19.5 grams. If the speed was measured to be 253 m/s, determine the tension to which it was pulled.

## Problem 14:

In a demonstration, Mr. H stretches a steel wire to a length 1.23 meters and braces both ends so that they are not free to vibrate. He attaches a fancy piece of equipment which he calls a mechanical oscillator to the wire and explains how it works. Then Mr. H turns the oscillator on and tunes the frequency to 588 Hz. To the amazement of the class, the wire begins vibrating in the sixth harmonic wave pattern. a. Determine the speed of waves within the wire. b. Determine the frequency at which the wire will vibrate with the first harmonic wave pattern. c. Determine the frequency at which the wire will vibrate with the second harmonic wave pattern.

a. 241 m/s b. 98 Hz c. 196 Hz

## Problem 15:

Olivia and Mason are doing a lab which involves stretching an elastic cord between two poles which are 98 cm apart. They use a mechanical oscillator to force the cord to vibrate with the third harmonic wave pattern when the frequency is 84 Hz. Determine the speed of vibrations within the elastic cord.

## Problem 16:

A 1.65-meter length string is forced to vibrate in its fifth harmonic. Determine the locations of the nodal positions. Express the locations as a distance measured from one of the ends of the string.

## Problem 17:

A steel piano wire is pulled to a tension of 448 N and has a mass density of 0.00621 kg/m. The string is 61.8 cm long and vibrates at its fundamental frequency. a. Determine the speed at which vibrations travel through the wire. b. Determine the wavelength of the standing wave pattern for the fundamental frequency. c. Determine the frequency of its vibrations.

a. 269 m/s b. 124 cm or 1.24 m c. 217 Hz

## Problem 18:

A steel piano wire is 72.9 cm long and has a mass of 4.54 x 10 -3 kg. The fundamental frequency of the wire is 262 Hz, corresponding to the frequency of middle C on the musical scales. Determine the tension to which the wire is pulled in order to vibrate with this frequency.

## Problem 19:

In the singing rod demonstration, Mr. H holds a 2.13-m length of aluminum rod in the exact center. With rosin on his fingers, he slides his sticky fingers back and forth over the rod until it begins to sing out with a very pure tone. Assuming waves travel at speeds of 6320 m/s within the aluminum and that the standing wave pattern is characteristic of one-half wavelength between the rod's ends, determine the frequency of the sound.

## Problem 20:

Two strings made of the same material (same mass density) and stretched to the same tension (and thus, having the same speed) have a different length. One of the strings is 80-cm long ( String A ) and the other string is 60 cm ( String B ). They are vibrated at various frequencies in order to establish standing wave patterns within them. Consider the first six harmonics of the two strings. Which harmonic of String A would have the same frequency as one of the harmonics of String B?

## Problem 21:

An 80-cm length open-end air column is forced to vibrate in its fifth harmonic. Determine the locations of the nodal positions (positions where air is undisturbed). Express the locations in cm using the diagram below.

## Problem 22:

An 80-cm length closed-end air column is forced to vibrate in its fifth harmonic. Determine the locations of the nodal positions (positions where air is undisturbed). Express the locations in cm using the diagram below. Note that the closed end is at 0 cm.

## Problem 23:

An organ pipe which acts as a open-end resonator has a length of 83 cm. Its fundamental frequency is 210 Hz. a. Determine the speed of sound waves in the air column of the pipe. b. Determine the frequency of the second, third and fourth harmonics of the organ pipe.

a. 350 m/s (rounded from 349 m/s) b. 420 Hz, 630 Hz, 840 Hz

## Problem 24:

An organ pipe which acts as a closed-end resonator has a length of 83 cm. Sound waves travel at 350 m/s through the air column of the pipe. a. Determine the fundamental frequency of the organ pipe. b. Determine the frequency of the next two harmonics of this closed-end organ pipe.

a. 110 Hz (rounded from 105.4 Hz) b. 320 Hz (rounded from 316 Hz) and 530 Hz (rounded from 527 Hz)

## Problem 25:

Middle C on a piano keyboard corresponds to a frequency of 262 Hz. Determine the length of an open-end air column whose fundamental matches this frequency. Use a value of 348 m/s as the speed of sound.

## Problem 26:

In the Speed of Sound Lab, Anna Litical and Noah Formula partially submerge a plastic tube into a column of water. The water air interface causes the tube to be a closed-end air column. Holding a 384-Hz tuning fork above the tube, they find that they must adjust the length of air within the air column to 22.6 cm in order to force the air column into resonance vibration with the tuning fork. Determine the speed of sound in the air column.

## Problem 27:

A musical recorder acts as an open-end air column, with a vibrational antinode located at the hole near the mouthpiece (known as the windway hole) and a vibrational antinode located at the nearest open tone hole. Blowing gently on the mouthpiece will force the air column to vibrate at its fundamental frequency. Assuming a speed of sound of 345 m/s, what length of air would be required to cause the recorder to sound out at 1050 Hz?

## Problem 28:

A 2.29-m long organ pipe acts as a closed-end resonator that produces several different harmonic frequencies in the audible range from 20 Hz to 20,000 Hz. Assuming a speed of sound of 343 m/s, determine the 5th highest frequency that the pipe can produce.

## Problem 29:

The fundamental frequency of an open-end organ pipe is 392 Hz. The third harmonic of a closed-end organ pipe has the same frequency. The speed of sound in air is 346 m/s. a. Determine the length of the open-end pipe. b. Determine the length of the closed-end pipe.

a. 0.441 m or 44.1 cm b. 0.662 m or 66.2 cm

## Problem 30:

Haley and Ariel are working on their Musical Instruments Project. They are creating a Pop Bottle Orchestra using a collection of pop bottles filled with water to varying heights. By blowing over the top of the pop bottles, the air column inside acts as a closed-end air column with the water surface at the bottom forcing a vibrational node within the standing wave pattern. Each bottle is 34.2-cm tall. Assuming a speed of sound of 345 m/s and a first harmonic wave pattern, determine how much water must be in the bottle in order for it to produce a 416 Hz sound wave.

## Problem 31:

South's orchestra walk on stage, take their seats and attempt to tune all instruments to 440 Hz - the A note of the oboe. Elizabeth tunes the A-string of her violin to the oboe. At one point in the process, she detects beats of 2 Hz between her string and the oboe. If her string is playing too low of a pitch, then what is its frequency?

## Problem 32:

Marc is attempting to tune his guitar. After adjusting his E4 string to 330 Hz, he begins to tackle the B3 string. In the process, he plucks both strings simultaneously and notices beats occurring at a rate of 12 beats in 4 seconds. When plucked individually, Marc observes the B3 string to sound out at a higher pitch. What is the frequency of the B3 string?

Return to Overview

## View Audio Guided Solution for Problem:

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## Heavy Machinery Meets AI

- Vijay Govindarajan
- Venkat Venkatraman

Until recently most incumbent industrial companies didn’t use highly advanced software in their products. But now the sector’s leaders have begun applying generative AI and machine learning to all kinds of data—including text, 3D images, video, and sound—to create complex, innovative designs and solve customer problems with unprecedented speed.

Success involves much more than installing computers in products, however. It requires fusion strategies, which join what manufacturers do best—creating physical products—with what digital firms do best: mining giant data sets for critical insights. There are four kinds of fusion strategies: Fusion products, like smart glass, are designed from scratch to collect and leverage information on product use in real time. Fusion services, like Rolls-Royce’s service for increasing the fuel efficiency of aircraft, deliver immediate customized recommendations from AI. Fusion systems, like Honeywell’s for building management, integrate machines from multiple suppliers in ways that enhance them all. And fusion solutions, such as Deere’s for increasing yields for farmers, combine products, services, and systems with partner companies’ innovations in ways that greatly improve customers’ performance.

Combining digital and analog machines will upend industrial companies.

## Idea in Brief

The problem.

Until recently most incumbent industrial companies didn’t use the most advanced software in their products. But competitors that can extract complex designs, insights, and trends using generative AI have emerged to challenge them.

## The Solution

Industrial companies must develop strategies that fuse what they do best—creating physical products—with what digital companies do best: using data and AI to parse enormous, interconnected data sets and develop innovative insights.

## The Changes Required

Companies will have to reimagine analog products and services as digitally enabled offerings, learn to create new value from data generated by the combination of physical and digital assets, and partner with other companies to create ecosystems with an unwavering focus on helping customers solve problems.

For more than 187 years, Deere & Company has simplified farmwork. From the advent of the first self-scouring plow, in 1837, to the launch of its first fully self-driving tractor, in 2022, the company has built advanced industrial technology. The See & Spray is an excellent contemporary example. The automated weed killer features a self-propelled, 120-foot carbon-fiber boom lined with 36 cameras capable of scanning 2,100 square feet per second. Powered by 10 onboard vision-processing units handling almost four gigabytes of data per second, the system uses AI and deep learning to distinguish crops from weeds. Once a weed is identified, a command is sent to spray and kill it. The machine moves through a field at 12 miles per hour without stopping. Manual labor would be more expensive, more time-consuming, and less reliable than the See & Spray. By fusing computer hardware and software with industrial machinery, it has helped farmers decrease their use of herbicide by more than two-thirds and exponentially increase productivity.

- Vijay Govindarajan is the Coxe Distinguished Professor at Dartmouth College’s Tuck School of Business, an executive fellow at Harvard Business School, and faculty partner at the Silicon Valley incubator Mach 49. He is a New York Times and Wall Street Journal bestselling author. His latest book is Fusion Strategy: How Real-Time Data and AI Will Power the Industrial Future . His Harvard Business Review articles “ Engineering Reverse Innovations ” and “ Stop the Innovation Wars ” won McKinsey Awards for best article published in HBR. His HBR articles “ How GE Is Disrupting Itself ” and “ The CEO’s Role in Business Model Reinvention ” are HBR all-time top-50 bestsellers. Follow him on LinkedIn . vgovindarajan
- Venkat Venkatraman is the David J. McGrath Professor at Boston University’s Questrom School of Business, where he is a member of both the information systems and strategy and innovation departments. His current research focuses on how companies develop winning digital strategies. His latest book is Fusion Strategy: How Real-Time Data and AI Will Power the Industrial Future. Follow him on LinkedIn . NVenkatraman

## IMAGES

## VIDEO

## COMMENTS

Solving Speed Problems. Flashcards. Learn. Test. Match. Flashcards. Learn. Test. Match. Created by. ... Draw a diagram to help you solve the problem. Verified answer. MATH. How many subsets does each of the following sets contain? (A) {a} (B) {a,b} (C) {a,b,c} (D) {a,b,c,} ... Other Quizlet sets. Business Law: Test 3 Study Guide. 58 terms ...

b. second and fifth hour. What is the average speed of the car for the first three hours of the trip? (2 points) b. 20 miles per hour. If Karen and her mother take three hours to come home, calculate the average speed of the car for the trip home. (2 points) c. 25 miles per hour. 10/10 Learn with flashcards, games, and more — for free.

A trip to Florida takes 10 hours. The distance is 816 km. Calculate the average speed. 3.75 sec. How many seconds will it take for a satellite to travel 450 m at a rate of 120 m/s ? 200 mi/hr. A plane travels 1000 miles in 5 hours. What is the plane's average speed? 10 m/s. A girl on a bike rides down a hill 500 meters long in 50 seconds.

Calculate the speed of a car that travels 556 kilometers northeast in 3.4 hours. Leave your answer in kilometers per hour. If the distance covered by a jogger is 2,541meters through the park and the time it took to cover that distance was 43.6 minutes, what was the speed of the jogger? Which object has a greater speed, a ball rolling down a 3.4 ...

Study with Quizlet and memorize flashcards containing terms like Yuto and Lian are at train stations 1,880 kilometers apart. Yuto boards a train heading east at an average speed of 220 kilometers per hour. At the same time, Lian boards a train heading west on a parallel track at an average speed of 250 kilometers per hour. How far has Lian traveled when the two trains pass each other?, Eduardo ...

0.97 m/s. Which object has a greater velocity, a ball rolling down a 3.4 meter hill in six seconds or a fish swimming upstream and covering 5.4 meters in 24 seconds. The ball, 0.57 m/s over 0.23m/s for the fish. Vocabulary and practice problems regarding velocity and speed. Learn with flashcards, games, and more — for free.

Study with Quizlet and memorize flashcards containing terms like Talia took the bus from her home to the bank and then walked back to her home along the same route. The round trip took 0.9 hours total. The bus traveled at an average speed of 40 km/h and she walked at an average speed of 5 km/h. Use the table to complete these statements. The rate of Trip 2 is ____ km/h. The time of Trip 1 is ...

Average speed is a scalar, so we do not include direction in the answer. We can check the reasonableness of the answer by estimating: 5 meters divided by 2 seconds is 2.5 m/s. Since 2.5 m/s is close to 2.9 m/s, the answer is reasonable. This is about the speed of a brisk walk, so it also makes sense.

Speed, velocity, and acceleration problems with detailed answers are provided for high school physics. Speed, velocity, and acceleration problems with detailed answers are provided for high school physics. ... By practicing these problems, you can get better at solving tough physics questions. This can help you do well in your exams. So, come ...

Solve coin word problems. Step 1. Read the problem. Make sure all the words and ideas are understood. Determine the types of coins involved. Create a table to organize the information. Label the columns "type," "number," "value," and "total value.". List the types of coins. Write in the value of each type of coin.

Distance, rate and time problems are a standard application of linear equations. When solving these problems, use the relationship rate (speed or velocity) times time equals distance. r⋅t = d r ⋅ t = d. For example, suppose a person were to travel 30 km/h for 4 h. To find the total distance, multiply rate times time or (30km/h) (4h) = 120 km.

Speed and Velocity. Speed is how fast something moves. Velocity is speed with a direction. Saying Ariel the Dog runs at 9 km/h (kilometers per hour) is a speed. But saying he runs 9 km/h Westwards is a velocity. Imagine something moving back and forth very fast: it has a high speed, but a low (or zero) velocity.

Solution: Step 1: The formula for distance is. Distance = Rate × Time. Total distance = 50 × 3 + 60 × 2 = 270. Step 2: Total time = 3 + 2 = 5. Step 3: Using the formula: Answer: The average speed is 54 miles per hour. Be careful! You will get the wrong answer if you add the two speeds and divide the answer by two.

Since the speed of the express train is 12 mph faster, we represent that as r+12. r r + 12 = speed of the local train = speed of the express train r = speed of the local train r + 12 = speed of the express train. Fill in the speeds into the chart. Now solve this equation. Find the speed of the express train.

Problem 5: If I can walk at an average speed of 5 km/h, how many miles I can walk in two hours? Solution to Problem 5: distance = (average speed) * (time) = 5 km/h * 2 hours = 10 km using the rate of conversion 0.62 miles per km, the distance in miles is given by distance = 10 km * 0.62 miles/km = 6.2 miles Problem 6: A train travels along a straight line at a constant speed of 60 mi/h for a ...

A useful problem-solving strategy was presented for use with these equations and two examples were given that illustrated the use of the strategy. Then, the application of the kinematic equations and the problem-solving strategy to free-fall motion was discussed and illustrated. In this part of Lesson 6, several sample problems will be presented.

Videos, worksheets, solutions and activities to help Algebra 1 students learn how to solve wind and current word problems. Solves this rate of wind problem using 2 variables and 2 linear equations. A plane flying against the wind flew 270 miles in 3 hours. Flying with the wind, the plane traveled 260 miles in 2 hours.

150 km × t 150 km = 250km × 3hours 150 km. Simplify. 150 km × t 150 km = 250km × 3hours 150 km. t = 250 × 3 150 hours = 5 hours. The exercises below with solutions and explanations are all about solving rate problems. Solve the following rate problems. The distance between two cities on the map is 15 centimeters.

Solution: This is a basic kinematics problem, so we will explain the steps in detail. Step 1: Since all these problems are in one dimension, draw a horizontal axis (like the positive x x axis), and place the object on it, so that its motion matches the direction of the axis. Step 2: Specify the known and wanted information.

Although speed and velocity are often words used interchangeably, in physics, they are distinct concepts. Velocity (v) is a vector quantity that measures displacement (or change in position, Δs) over the change in time (Δt), represented by the equation v = Δs/Δt. Speed (or rate, r) is a scalar quantity that measures the distance traveled (d ...

Find the distance between the two cities. Solution to Problem 7: Let x be John's rate in traveling between the two cities. The rate of Peter will be x + 10. We use the rate-time-distance formula to write the distance D traveled by John and Peter (same distance D) D = 3 x and D = 2 (x + 20) The first equation can be solved for x to give.

You might need: Calculator. A rocket ship starts from rest and turns on its forward booster rockets, causing it to have a constant acceleration of 4 m s 2 rightward. After 3 s , what will be the velocity of the rocket ship? Answer using a coordinate system where rightward is positive. m s.

Problem 24: An organ pipe which acts as a closed-end resonator has a length of 83 cm. Sound waves travel at 350 m/s through the air column of the pipe. a. Determine the fundamental frequency of the organ pipe. b. Determine the frequency of the next two harmonics of this closed-end organ pipe. Audio Guided Solution.

The automated weed killer features a self-propelled, 120-foot carbon-fiber boom lined with 36 cameras capable of scanning 2,100 square feet per second. Powered by 10 onboard vision-processing ...