Applied Problems: Motion

By Dr. Marcia L.Tharp

Dr. Julia Arnold

Applied Problems: Motion

Motion problems use the equation

D = RT

where D is the distance traveled.

R is the rate of travel.

T is the time spent traveling .

Applied Problems: Motion

Motion Problem 1 Let’s suppose that you drive your car 120 miles in 2 hours than we can find the rate you traveled by using the distance equation.

D = RT

Since 120 is a distance we let D = 120 Since the time you traveled is 2 hours we let T= 2 .

Placing this in the equation D=RT we have 120 = R 2 120 = R 2 2 2 60 = R

So your rate is 60 miles per hour.

Motion Problem 2 Pat has been caught speeding by the airplane patrol. He is doing 85miles per hour. A police car who is 30 miles behind him can do 100 miles per hour to catch him. How long will it take the police car to catch Pat if they continue in the same direction at this speed?

Print this page.

It is helpful to use a

each distance.

D = RT grid

when solving motion problems as shown in the following example.

The purpose of the grid is to find an algebraic name for

D=RT Grid

To use this grid fill in the rate and time for each driver.

We will let x = the time it takes for the police car to catch Pat. Then multiply the rate by the time to get the distance traveled.

Rate Time Distance Pat Police Car

Pat Police Car

D=RT Grid

Rate Time Distance 85 100

D=RT Grid

Rate Time Distance Pat 85 Police Car 100 X X

D=RT Grid

Rate Time Distance Pat 85 X 85 x Police Car 100 X 100 x

Notice that neither the policeman or Pat are represented by the 30 miles. We will use this later.

Next Draw A Picture

Picture

30 miles Pat’s distance

Police Car’s distance

Draw A Picture

Picture

30 miles Pat’s distance

Police Car’s distance

Use the picture to write an equation.

30 miles Pat’s distance Police Car’s distance Police Car’s = 30 miles + Pat’s Distance Distance

Now fill in the equation using our chart.

30 miles Pat’s distance Police Car’s distance Police Car’s = 30 miles + Pat’s Distance Distance

Rate Time Distance 100x = 30 + 85x Pat 85 X Police Car 100 X 85x 100x

Next solve this equation.

Subtract 85 x from both sides.

Divide by 15 on each side.

100x = 30 + 85x 100x – 85x = 30 + 85x – 15x = 30 85x 15x = 30 15 15 x = 2

hours

So it took the police car 2 hours to catch Pat.

Problem 3 Juan and Amal leave DC at the same time headed south on I-95. If Juan averages 60 mph and Amal averages 72 mph how long will it take them to be 30 miles apart? (Now would be a good time for a guess. Write yours down and try it in this table.) Juan Amal Rate 60 72 Time x x Distance 60x 72x

Juan Amal Rate 60 72 Time x x Distance 60x 72x

We are looking for a time where Juan and Amal will be 30 miles apart. How do we represent the distance between the two men?

60x - 72x Or is it 72x - 60x . Which of these two would be positive?

The correct equation is 72x - 60 x = 30

Sherry and Bob like to jog in the park. Sherry can jog at 5 mph, while Bob can jog at 7 mph.

If Sherry starts 30 minutes ahead of Bob, how long will it take Bob to catch up to Sherry?

Draw a grid like the one below and fill in what you know about Sherry and Bob.

Rate Time Distance Sherry Bob

When finished go to the next slide to see how you did.

Sherry and Bob like to jog in the park. Sherry can jog at 5 mph, while Bob can jog at 7 mph.

If Sherry starts 30 minutes ahead of Bob, how long will it take Bob to catch up to Sherry?

Let x = the time it takes Bob to catch Sherry

Rate Time Distance Sherry Bob

5 7 x + 1/2 x 5(x + 1/2) 7x Sherry started 30 minutes or 1/2 hour before Bob, so her time must reflect that amount. Rate is in miles per hour, time must be in hours so our units match.

Sherry’s jogging distance Sherry Bob The problem is finished when Bob catches up to Sherry.

Thus, their distances must equal each other.

5(x + 1/2) = 7x 5x + 5/2 = 7x 5/2 = 2x 2[5/2] = 2 (2x) 5 = 4x x = 5/4 = 1.25 hrs

Now it’s time to see what You can do.

Practice Problems: Motion 1. Tonya and Freda drive away from Norfolk on the same road in the same direction. If Tonya is averaging 52 mph and Freda is averaging 65 mph, how long will it take for them to be 39 miles apart? 2. Tonya and Freda drive away from Norfolk on the same road in opposite directions. If Tonya is averaging 52 mph and Freda is averaging 65 mph, how long will it take for them to be 39 miles apart? Round your answer to the nearest minute.

3. Bernadette drove 120 miles. The first part of the trip she averaged 60 mph, but on the second part of the trip she ran into some congestion and averaged 48 mph. If the total driving time was 2.2 hours, how much time did she spend at 60 mph? 4. Dr. John left New Orleans at 12 noon. His drummer left at 1:00 traveling 9 mph faster. If the drummer passed Dr. John at 6:00, what was the average speed of each?

Motion problem solutions: 1.

3 hours 2.

1/3 hour = 20 minutes 3.

1.2 hours at 60 mph 4.

45 mph for Dr. John and 54 mph for the drummer For worked out solutions click to next slide.

1. Tonya and Freda drive away from Norfolk on the same road in the same direction. If Tonya is averaging 52 mph and Freda is averaging 65 mph, how long will it take for them to be 39 miles apart? Let x = time it takes for them to be 39 miles apart.

Construct a table to put your information in.

Tonya and Freda drive away from Norfolk on the same road in the same direction. If Tonya is averaging 52 mph and Freda is averaging 65 mph, how long will it take for them to be 39 miles apart?

Tonya Freda Rate 52 65 Time x x Distance 52x 65x We let x stand for the time they have been driving which would be the same in this case. How far they have gone (distance) is written in the chart by using the known information and the formula R*T=D

Tonya and Freda drive away from Norfolk on the same road in the same direction. If Tonya is averaging 52 mph and Freda is averaging 65 mph, how long will it take for them to be 39 miles apart?

Tonya Freda Rate 52 65 Time x x Distance 52x 65x Tonya and Freda are headed in the same direction so we can picture their distances as this: Tonya 52x 39 mi Freda 65x Do you see the equation forming from our picture?

Tonya and Freda drive away from Norfolk on the same road in the same direction. If Tonya is averaging 52 mph and Freda is averaging 65 mph, how long will it take for them to be 39 miles apart?

Rate Time Distance Tonya Freda 52 65 x x 52x 65x Tonya Freda 52x 65x 39 mi One way to look at it might be to say 52x + 39 = 65x Or another way might be to say 65x – 52x = 39 Both are correct and will give you the correct answer of 3 hrs.

2. Tonya and Freda drive away from Norfolk on the same road in opposite directions. If Tonya is averaging 52 mph and Freda is averaging 65 mph, how long will it take for them to be 39 miles apart? Round your answer to the nearest minute.

Now they are going in opposite directions, but we will begin the same way by constructing our table.

Tonya and Freda drive away from Norfolk on the same road in opposite directions. If Tonya is averaging 52 mph and Freda is averaging 65 mph, how long will it take for them to be 39 miles apart? Round your answer to the nearest minute.

Tonya Rate 52 Time x Distance 52x Freda 65 x 65x As you can see the table is the same, so only the picture of the event must change. Now it looks like this: S T A R T 52x 65x Tonya Freda 39 miles Do you see the equation forming from this picture?

Tonya and Freda drive away from Norfolk on the same road in opposite directions. If Tonya is averaging 52 mph and Freda is averaging 65 mph, how long will it take for them to be 39 miles apart? Round your answer to the nearest minute.

Tonya Rate 52 Time x Distance 52x Freda 65 x 65x S T A R T 52x 65x Tonya 39 miles 52x + 65x = 39 117x=39 X= 1/3 of an hour or 1/3 of 60 min = 20 minutes Freda

3. Bernadette drove 120 miles. The first part of the trip she averaged 60 mph, but on the second part of the trip she ran into some congestion and averaged 48 mph. If the total driving time was 2.2 hours, how much time did she spend at 60 mph? The question here deals with time. Again, lets Fill in the chart.

Bernadette drove 120 miles. The first part of the trip she averaged 60 mph, but on the second part of the trip she ran into some congestion and averaged 48 mph. If the total driving time was 2.2 hours, how much time did she spend at 60 mph?

1 st leg Rate 60 Time t Distance 60t 2 nd leg 48 2.2 - t 48(2.2 – t) This time totals are given for both time traveled and distance traveled. Thus totals don’t belong in the chart. Bernadette did not travel 2.2 hours at 60 mph nor did she travel 2.2 hours at 48 mph. She traveled 2.2 hours total at both of those speeds. So how do we write this in the chart. Click to observe.

Distance is again computed by formula R*T = D

Bernadette drove 120 miles. The first part of the trip she averaged 60 mph, but on the second part of the trip she ran into some congestion and averaged 48 mph. If the total driving time was 2.2 hours, how much time did she spend at 60 mph?

1 st leg 2 nd leg Rate 60 48 Time t 2.2 - t Distance 60t 48(2.2 – t) What is the picture for this problem?

60t 48(2.2 – t) Dist 1 st leg Dist 2 nd leg Total dist. 120 miles Do you see the equation from the picture?

60t +48(2.2 – t) = 120

Bernadette drove 120 miles. The first part of the trip she averaged 60 mph, but on the second part of the trip she ran into some congestion and averaged 48 mph. If the total driving time was 2.2 hours, how much time did she spend at 60 mph? 60t 48(2.2 – t) Dist 1 st leg Dist 2 nd leg Total dist. 120 miles 60t +48(2.2 – t) = 120 60t + 105.6 – 48t = 120 12t = 14.4

t = 1.2 hours at 60mph

4. Dr. John left New Orleans at 12 noon. His drummer left at 1:00 traveling 9 mph faster. If the drummer passed Dr. John at 6:00, what was the average speed of each?

Now we don’t know the speed. We do know something about time. Let’s see how we can fill in the chart.

Dr. John left New Orleans at 12 noon. His drummer left at 1:00 traveling 9 mph faster. If the drummer passed Dr. John at 6:00, what was the average speed of each? John Rate x Time 6 Distance 6x Drumme X + 9 5 5(x + 9) r John at 6 PM, so how long has Dr. John been driving?

How long has the drummer been driving?

What is our picture?

Dr. John 6x Drummer 5x + 45 Drummer passing Dr. John

Dr. John left New Orleans at 12 noon. His drummer left at 1:00 traveling 9 mph faster. If the drummer passed Dr. John at 6:00, what was the average speed of each? John Rate x Time 6 Distance 6x Drumme r Dr. John X + 9 5 5(x + 9) Drummer What equation does the picture suggest?

The two distances are equal thus: 6x = 5(x + 9) 6x = 5x + 45 X = 45 mph for Dr. John X + 9 = 54 mph for the drummer

Now go on to the Geometry Lessons.

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