Displacement given initial velocity, time, and acceleration: delta s = v0 t + (1/2) a t^2. If v0 = 3 m/s, a = 2 m/s^2, t = 5 s, what is delta s?

Displacement given initial velocity, time, and acceleration: delta s = v0 t + (1/2) a t^2. If v0 = 3 m/s, a = 2 m/s^2, t = 5 s, what is delta s?

• A. 50 m
• B. 25 m
• C. 15 m
• D. 40 m

Answer: (D)

Displacement under constant acceleration comes from two contributions: distance from the initial velocity over the time interval, and the extra distance gained as velocity increases due to acceleration. Use delta s = v0 t + (1/2) a t^2. Plug in the values: - v0 t = 3 m/s × 5 s = 15 m - (1/2) a t^2 = 0.5 × 2 m/s^2 × (5 s)^2 = 0.5 × 2 × 25 = 25 m Add them: 15 m + 25 m = 40 m. The displacement is 40 meters. Common missteps: dropping the initial-velocity term gives 25 m, dropping the acceleration term gives 15 m, or using a t^2 instead of (1/2) a t^2 gives 50 m. The correct expression with both parts and the 1/2 factor yields 40 m.

Displacement under constant acceleration comes from two contributions: distance from the initial velocity over the time interval, and the extra distance gained as velocity increases due to acceleration. Use delta s = v0 t + (1/2) a t^2.

Plug in the values:

• v0 t = 3 m/s × 5 s = 15 m

• (1/2) a t^2 = 0.5 × 2 m/s^2 × (5 s)^2 = 0.5 × 2 × 25 = 25 m

Add them: 15 m + 25 m = 40 m. The displacement is 40 meters.

Common missteps: dropping the initial-velocity term gives 25 m, dropping the acceleration term gives 15 m, or using a t^2 instead of (1/2) a t^2 gives 50 m. The correct expression with both parts and the 1/2 factor yields 40 m.