A velocity-time graph shows a straight line from t = 0 to t = 4 s with velocity increasing from 0 to 20 m/s. What are the acceleration and the displacement during this interval?

• A. a = 5 m/s^2; displacement = 40 m
• B. a = 4 m/s^2; displacement = 40 m
• C. a = 5 m/s^2; displacement = 20 m
• D. a = 5 m/s^2; displacement = 80 m

Answer: (A)

This question checks how a velocity-time graph encodes acceleration and displacement. The slope of a velocity-time graph equals acceleration, and the area under the curve gives displacement.

From 0 to 4 seconds, velocity rises from 0 to 20 m/s. The acceleration is Δv/Δt = (20 − 0) / (4 − 0) = 5 m/s^2.

The displacement is the area under the velocity-time graph during that interval. The line forms a triangle with base 4 s and height 20 m/s, so area = 1/2 × base × height = 1/2 × 4 × 20 = 40 m.

So the acceleration is 5 m/s^2 and the displacement is 40 m. If a different value were chosen, it would imply a different Δv or a different area under the curve.