The displacement during the 5-second interval in which velocity rises from 0 to 20 m/s is?

The displacement during the 5-second interval in which velocity rises from 0 to 20 m/s is?

• A. 100 m
• B. 50 m
• C. 150 m
• D. 0 m

Answer: (B)

Displacement equals the area under the velocity-time graph, since velocity is how position changes over time. Here velocity increases linearly from 0 to 20 m/s over 5 seconds, so the v–t graph is a triangle with base 5 s and height 20 m/s. The area is (1/2) × base × height = 0.5 × 5 × 20 = 50 m. You can also see this via constant acceleration: a = Δv/Δt = 20/5 = 4 m/s², and displacement s = (1/2) a t² = (1/2) × 4 × (5)² = 50 m. The other numbers don’t match the area under the curve or the kinematics for this interval.

Displacement equals the area under the velocity-time graph, since velocity is how position changes over time. Here velocity increases linearly from 0 to 20 m/s over 5 seconds, so the v–t graph is a triangle with base 5 s and height 20 m/s. The area is (1/2) × base × height = 0.5 × 5 × 20 = 50 m.

You can also see this via constant acceleration: a = Δv/Δt = 20/5 = 4 m/s², and displacement s = (1/2) a t² = (1/2) × 4 × (5)² = 50 m.

The other numbers don’t match the area under the curve or the kinematics for this interval.