A velocity-time graph shows velocity increasing linearly from 0 to 20 m/s over 5 s. What distance is traveled in those 5 s?

A velocity-time graph shows velocity increasing linearly from 0 to 20 m/s over 5 s. What distance is traveled in those 5 s?

• A. 50 m
• B. 100 m
• C. 25 m
• D. 75 m

Answer: (A)

Distance traveled is the area under a velocity-time graph. Here velocity increases linearly from 0 to 20 m/s over 5 seconds, so the graph is a triangle with base 5 s and height 20 m/s. The area is (1/2) × 5 × 20 = 50 m, which is the distance traveled. Another way to see it is that the motion has constant acceleration, so the average velocity is (0 + 20)/2 = 10 m/s; multiplying by the 5 s gives 50 m. The other options would require a larger or smaller average velocity than 10 m/s, which isn’t the case in this situation.

Distance traveled is the area under a velocity-time graph. Here velocity increases linearly from 0 to 20 m/s over 5 seconds, so the graph is a triangle with base 5 s and height 20 m/s. The area is (1/2) × 5 × 20 = 50 m, which is the distance traveled. Another way to see it is that the motion has constant acceleration, so the average velocity is (0 + 20)/2 = 10 m/s; multiplying by the 5 s gives 50 m. The other options would require a larger or smaller average velocity than 10 m/s, which isn’t the case in this situation.