A velocity-time graph shows velocity increasing uniformly from 0 to 12 m/s over 6 s. What are the acceleration and the distance traveled in that interval?

• A. Acceleration 2 m/s^2; distance 30 m
• B. Acceleration 1 m/s^2; distance 36 m
• C. Acceleration 2 m/s^2; distance 39 m
• D. Acceleration 2 m/s^2; distance 36 m

Answer: (D)

A velocity-time graph with velocity increasing uniformly shows constant acceleration, and the slope of the line gives the acceleration. Here the velocity changes from 0 to 12 m/s in 6 s, so the acceleration is a = Δv/Δt = (12 − 0)/6 = 2 m/s^2.

The distance traveled is the area under the velocity-time graph over that interval. That graph forms a triangle with base 6 s and height 12 m/s, so distance = 1/2 × 6 × 12 = 36 m. You can also use s = v0 t + 1/2 a t^2: s = 0 + 1/2 × 2 × (6)^2 = 36 m.

So the acceleration is 2 m/s^2 and the distance traveled is 36 m.