Displacement given final velocity, acceleration, and time: delta s = vf t - (1/2) a t^2. If vf = 7 m/s, a = 2 m/s^2, t = 3 s, what is delta s?

Displacement given final velocity, acceleration, and time: delta s = vf t - (1/2) a t^2. If vf = 7 m/s, a = 2 m/s^2, t = 3 s, what is delta s?

• A. 9 m
• B. 12 m
• C. 18 m
• D. 21 m

Answer: (B)

With constant acceleration, displacement over a time t can be written in terms of the final velocity: delta s = v_f t − (1/2) a t^2. This comes from starting with v_f = v0 + a t and s = v0 t + (1/2) a t^2, solving for v0 and substituting to eliminate it. Plug in the numbers: v_f t = 7 m/s × 3 s = 21 m. The second term is (1/2) a t^2 = 0.5 × 2 m/s^2 × (3 s)^2 = 1 × 9 = 9 m. Subtracting gives delta s = 21 m − 9 m = 12 m. As a quick consistency check, the initial velocity would be v0 = v_f − a t = 7 − 2×3 = 1 m/s, so the average velocity is (v0 + v_f)/2 = (1 + 7)/2 = 4 m/s, and times the 3 s interval yields 12 m, matching the result.

With constant acceleration, displacement over a time t can be written in terms of the final velocity: delta s = v_f t − (1/2) a t^2. This comes from starting with v_f = v0 + a t and s = v0 t + (1/2) a t^2, solving for v0 and substituting to eliminate it.

Plug in the numbers: v_f t = 7 m/s × 3 s = 21 m. The second term is (1/2) a t^2 = 0.5 × 2 m/s^2 × (3 s)^2 = 1 × 9 = 9 m. Subtracting gives delta s = 21 m − 9 m = 12 m.

As a quick consistency check, the initial velocity would be v0 = v_f − a t = 7 − 2×3 = 1 m/s, so the average velocity is (v0 + v_f)/2 = (1 + 7)/2 = 4 m/s, and times the 3 s interval yields 12 m, matching the result.