A cyclist pedaling increases speed from 5 m/s to 9 m/s over 4 s. What is the average acceleration, and what distance is covered during this interval if the initial velocity is 5 m/s?

• A. a = 0.5 m/s^2; s = 20 m
• B. a = 2 m/s^2; s = 40 m
• C. a = 1 m/s^2; s = 28 m
• D. a = 1.5 m/s^2; s = 35 m

Answer: (C)

The question tests how to find average acceleration and the distance covered with constant acceleration. Average acceleration is the change in velocity over time: ā = Δv/Δt. Here Δv = 9 m/s − 5 m/s = 4 m/s and Δt = 4 s, so ā = 4/4 = 1 m/s^2.

To get the distance, use s = ut + 0.5 a t^2 with initial speed u = 5 m/s, time t = 4 s, and a = 1 m/s^2. This gives s = 5×4 + 0.5×1×16 = 20 + 8 = 28 m. (You can also use the average speed approach: average speed is (5 + 9)/2 = 7 m/s, and distance is 7 × 4 = 28 m.)

The correct values are acceleration 1 m/s^2 and distance 28 m.