Question

Two cars A and B are at rest at the same point. Car A starts moving with a constant velocity of 40 m/s. Car B starts at the same time and moves in the same direction with a constant acceleration of 4 m/s².

• A. 800 m
• B. 400 m
• C. 1600 m
• D. 3200 m

Answer 1

Answer: (A)

800 m

Answer 2

Car A's position at time $t$: $S_A = 40t$. Car B's position at time $t$: $S_B = \frac{1}{2}(4)t^2 = 2t^2$. Overtaking occurs when $S_A = S_B$, so $40t = 2t^2$. Solving for $t$: $2t^2 - 40t = 0 \implies 2t(t-20) = 0$. The non-trivial time is $t=20$ s. The distance of overtaking is $S = 40 \times 20 = 800$ m.