The drawing shows velocity (v) versus time (t) 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 cysclist meet ?

Options

A. 4 min
B. 6 min
C. 8 min
D. 12 min

Answer

B

Explanation

`tan theta_1 = (v_0)/(4), tan theta_2 = (v_0)/(1)`
`rArr 4 tan theta_1 = tan theta_2 = v_0`
Let they meet at time `t`, then their displacement should be same. It means area under `v-t` graph should be same.
`S_1 = S_2`
`rArr (1)/(2) v_1 t = (1)/(2) v_2 (t - 3)`
`rArr (1)/(2)[(tan theta_1)t] t = (1)/(2)[(tan theta_2)(t - 3)](t - 3)`
`rArr (tan theta_1)t^2 = 4 (tan theta_1)(t - 3)^2`
`rArr t = 2(t - 3) rArr t = 6 min`.