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. `sqrt(100)s`
B. `sqrt(120)s`
C. `10sqrt(2)S`
D. `15s`

Answer

c

Explanation

let the bus and the car meet after time t.
The distance travelled by the car is given by :
`S_(1) =(1)/(2) xx 4 xxt^(2) = 2t^(2)`
The distance travellled by the bus is given by :
`S_(2) = (1)/(2) xx 2 xx t^(2) = t^(2)`
`S_(1) - S_(2) = 2t^(2) - t^(2) = t^(2)`
`t^(2) = 200`
`t = sqrt (200) = 10sqrt(2) s`.