The distance *d* that an object will travel is equal to its rate *r* times its time *t*.

*d* = *r t*

Rate is sometimes called speed or velocity. Rates are given as ratios, such as “miles per hour” or mph and “kilometers per hour” or km/h.

The distance covered in 1 hour at 20 miles per hour is the same as the distance covered in 30 minutes at 40 miles per hour.

*d* = *r t*, so 20 mph × 1 hour = 40 mph × 30 minutes

When setting up an equation, make sure that the units match. The rates 1 mile per minute and 60 miles per hour are the same.

1 mile/1 minute × 60 minutes/1 hour = 60 miles/1 hour

The most common methods for solving motion problems are to use a system of equations or to use a proportion.

**Converting Rates and Speeds**

## Example

A biker traveled 60 miles in 2.5 hours. Find the biker’s average speed.

### Solution

*d *= *r t*

60 = *r* × 2.5

*r* = 60 / 2.5 = 24 Divide both sides by 2.5.

*r* = 24

The biker’s average speed was 24 miles per hour.

## Example

An airplane travels 1,200 miles in 2.5 hours. How far will it travel in 10 hours?

### Solution

Use a proportion.

1200 miles/2.5 hours = *m* miles/10 hours

*m* = 1200 × 10/2.5 hours = 1200 × 100/25 = 1200 × 4 = 4800

The plane will travel 4,800 miles in 10 hours.

## Example

Bill takes

xhours to runymiles. On Monday, Bill will run in a marathon that iszmiles long. How long will he take to finish?(A)

xy(B) \dfrac{\textit{y}}{\textit{z}}

(C) \dfrac{\textit{x}}{\textit{yz}}

(D) \dfrac{\textit{xz}}{\textit{y}}

(E) \dfrac{\textit{yz}}{\textit{x}}

### Solution

Use *d =* *r t* and a proportion. The question assumes the average rate will be the same for the practice and the marathon. So the ratio of *r* = *d* / *t* will be the same.

Taking *x* hours to run *y* miles gives time and distance. The marathon is *z* miles long, so that is the distance. You need to find the time it takes to run *z* miles.

*y* miles/*x* hours = *z* miles/? hours

? = *xz*/*y*

The correct answer choice is (D).

## Example

A police officer, traveling at 100 miles per hour, pursues Philip, who has a 30-minute head start. The police officer overtakes Philip in two hours. Find Philip’s speed.

### Solution

The distance traveled by the officer equals the distance traveled by Philip.

Let *r* miles per hour be Philip’s speed. Use *d* = *r t*.

distance of the police officer = 100 mph × 2 hours

distance of Philip = *r* mph × 2.5 hours. The head start added half an hour to Philip’s time.

100 mph × 2 hours = *r* mph × 2.5 hours

200 = 2.5*r*

*r* = 200 / 2.5 = 80

Philip was driving at 80 miles per hour.

## Example

It takes 7 hours for a car to drive the 400 miles from City V to City W. The return trip takes 9 hours. Find the average speed of the car.

### Solution

Using *r* = *d*/*t* , the average rate = total distance / total time

The total distance is 2(400) = 800 miles.

The total time is 7 + 9 = 16 hours.

*r* = 800/16 = 50

The average speed was 50 miles per hour.

## Example

Joe drove 2.5 hours at 60 miles per hour and half an hour at 35 miles per hour. What was Joe’s average speed?

### Solution

Using *d *= *r t*:

distance = (2.5 × 60) + (0.5 × 35) = 167.5 miles

time = 2.5 + 0.5 = 3 hours

So 167.5 = 3*r*, and *r* ≈ 56. His average speed was 56 miles per hour.

Notice that the average speed is not (60 + 35)/2 = 52.5

The average rate = total distance/total time

On the graph below, the points *A*, *B*, *C* and *D* are different towns. The dashed lines show the route of a car travelling from town *A* to town *B* to town *C* to town *D* and back to *A*. The graph also shows the average speeds of the car on each leg. The car made no stops in any of the towns.

## Example

If the driver wants to repeat his

ABCDAtrip, what must be his speed on legDAso that the trip would last 3 hours? Use the graph above. Assume that the speed on other legs remains unchanged.(A) 35 km/h

(B) 42 km/h

(C) 45 km/h

(D) 54 km/h

(E) 60 km/h

### Solution

Use *t* = *d* / *r*. The original journey was:

- 30 km at 90 km/h = 30 / 90 = 1/3 hour
- 50 km at 50 km/h = 50 / 50 = 1 hour (Note: The distance of 50 is easy to find since it is the hypotenuse of a 3 : 4 : 5 right triangle.)
- 60 km at 60 km/h = 60 / 60 = 1 hour
- 40 km at 40 km/h = 40 / 40 = 1 hour

So the time for the original journey was 1/3 + 1 + 1 + 1 = 3 1/3 hours

To complete the trip in 3 hours, the driver needs to take 1/3 hour less. Specifically, he needs to take 1/3 hour less on the last leg, *DA*.

*DA* took 1 hour, so it needs to take 1 − 1/3 = 2/3 of an hour.

For the time to decrease, the speed needs to increase. Use *r* = *d* /* t*.

40 km in 2/3 hour = *r* km/h

*r* = 40 / (2 / 3) = 40 × (3 / 2) = 60

He will need to drive the last leg at 60 km/h. The correct answer choice is (E).

## Example

On another

ABCDAtrip, the car went 25% slower on each leg. About how much longer did the trip take? Use the graph above.(A) 27 minutes

(B) 40 minutes

(C) 67 minutes

(D) 240 minutes

(E) 267 minutes

### Solution

Since the speed decreased, the time increased.

Since the speed was 25% slower, the new speed was 100% – 25% = 75% of the original speed.

Use *d* = *r t*. The distances are the same.

(original rate) × (original time) = (new rate) × (new time)

(original rate) × (original time) = (0.75 × original rate) × (new time)

original time = 0.75 × new time = 3/4 × new time

The original time was 200 minutes.

200 × 4/3 = new time

266.66 ≈ 267 = new time

Be careful. The question doesn’t ask for the new time, it asks for the difference between the old and new times. A common trick on the SAT is for the answer choices to include a previous calculation. Remember to read all the answer options before making your choice.

267 – 200 = 67 minutes

The correct answer choice is (C).

#### The Seven Steps to Rate/Distance Problems

**Before attempting these problems, be sure to review this section on Quantitative Comparison questions.**

https://www.youtube.com/watch?v=lyMFLDi13C0&list=PLD0D070C218D8F5A3&index=11

https://www.youtube.com/watch?v=5RRjeLoqYhg&list=PLD0D070C218D8F5A3&index=13