Which expression gives displacement when initial velocity, time, and acceleration are known?

Which expression gives displacement when initial velocity, time, and acceleration are known?

• A. Delta s = vf t - (1/2) a t^2
• B. Delta s = v0 t + (1/2) a t^2
• C. Delta s = (1/2) (vf + v0) t
• D. Delta s = distance traveled

Answer: (B)

For constant acceleration, displacement accumulates as velocity changes linearly with time. The velocity at any time is v(t) = v0 + a t, so the displacement over a interval t is the integral of velocity: Δs = ∫0^t (v0 + a t′) dt′ = v0 t + (1/2) a t^2. If you set the starting position to zero, this becomes Δs = v0 t + (1/2) a t^2. This form directly uses the known initial velocity, time, and acceleration. Note that an equivalent expression can be written using the final velocity vf, since vf = v0 + a t, giving Δs = vf t − (1/2) a t^2. They’re the same thing, just written in different forms. The version that explicitly uses the initial velocity is the standard choice when those quantities are given.

For constant acceleration, displacement accumulates as velocity changes linearly with time. The velocity at any time is v(t) = v0 + a t, so the displacement over a interval t is the integral of velocity: Δs = ∫0^t (v0 + a t′) dt′ = v0 t + (1/2) a t^2. If you set the starting position to zero, this becomes Δs = v0 t + (1/2) a t^2. This form directly uses the known initial velocity, time, and acceleration.

Note that an equivalent expression can be written using the final velocity vf, since vf = v0 + a t, giving Δs = vf t − (1/2) a t^2. They’re the same thing, just written in different forms. The version that explicitly uses the initial velocity is the standard choice when those quantities are given.