The drawing shows velocity (v) versus time (r) graphs for two cyclists moving along the same straight segment of a highway from the same point. The second cyclist starts moving at t = 3 min. At what time do the two cyclists meet?

tan⁡θ1=v04,tan⁡θ2=v01⇒ 4tan⁡θ1=tan⁡θ2=v0Let they meet at time t, then their displacement should be same.It means area under v-t graph should be same.S1=S2⇒12v1t=12v2(t−3)⇒12tan⁡θ1tt=12tan⁡θ2(t−3)(t−3)⇒tan⁡θ1t2=4tan⁡θ1(t−3)2⇒t=2(t−3)⇒t=6min