A car starts from rest with a constant acceleration of 3.0 m/s^2. What distance does it travel in the first 4.0 s?

A car starts from rest with a constant acceleration of 3.0 m/s^2. What distance does it travel in the first 4.0 s?

• A. 24 m
• B. 12 m
• C. 36 m
• D. 48 m

Answer: (A)

When motion starts from rest with constant acceleration, the distance traveled is given by d = v0 t + (1/2) a t^2. Because it starts from rest, v0 = 0, so d = (1/2) a t^2. With a = 3.0 m/s^2 and t = 4.0 s, d = 0.5 × 3.0 × (4.0)^2 = 1.5 × 16 = 24 m. So the car covers 24 meters in the first 4 seconds. The other distances would come from different times or a nonzero initial velocity, which isn’t the situation here.

When motion starts from rest with constant acceleration, the distance traveled is given by d = v0 t + (1/2) a t^2. Because it starts from rest, v0 = 0, so d = (1/2) a t^2. With a = 3.0 m/s^2 and t = 4.0 s, d = 0.5 × 3.0 × (4.0)^2 = 1.5 × 16 = 24 m. So the car covers 24 meters in the first 4 seconds. The other distances would come from different times or a nonzero initial velocity, which isn’t the situation here.