A car is standing 200m behind a bus , which is also at rest . The two. Start moving at the same instant but with different forward accelerations. The bus has acceleration `2 ms ^(-2)` and The car has acceleration `4 ms^(-2)` The car will catch up will the bus after time :

Options

A. `sqrt100s`
B. `sqrt120s`
C. `10 sqrt2s`
D. 15s

Answer

C

Explanation

Given `u_(C) = u_(B) = o, a_(C) = 4m//s^(2), a_(B) = 2m//s^(2)` hence relative acceleration, `a_(CB) = 2m//sec^(2)` Now, we know, `s= ut + (1)/(2) "at"^(2) " " 200 = (1)/(2) xx 2t^(2) :. u=0`
Hence, the car will catch up with the bus after time `t= 10 sqrt2` second.