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
A biker traveled 60 miles in 2.5 hours. Find the biker’s average speed.
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.
An airplane travels 1,200 miles in 2.5 hours. How far will it travel in 10 hours?
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.
Bill takes x hours to run y miles. On Monday, Bill will run in a marathon that is z miles long. How long will he take to finish?
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).
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.
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 = 2.5r
r = 200 / 2.5 = 80
Philip was driving at 80 miles per hour.
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.
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.
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?
Using d = r t:
distance = (2.5 × 60) + (0.5 × 35) = 167.5 miles
time = 2.5 + 0.5 = 3 hours
So 167.5 = 3r, 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.
If the driver wants to repeat his ABCDA trip, what must be his speed on leg DA so 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
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).
On another ABCDA trip, 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
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.