Initially car A is 10.5 m ahead of car B. Both start moving at t=0 in the same direction. The velocity-time graph is shown. The time when car B catches car A is:

Question

G1 Physics · Accepted Answer

Car A moves at 10 m s⁻¹, while car B has v=t from the 45° v-t line. B must gain the initial 10.5 m lead: ∫₀ᵗ(vᵦ-vₐ)dt = t²/2-10t =10.5. Thus t²-20t-21=0, giving t=21 s. Hence option A.