A man runs at a speed of `4m//s` to overtake a standing bus. When he is `6m` beind the door at `t=0`, the bus mover forward and continuous with a constant acceleration of `1.2m//s^(2)`. The man reaches the door in time `t`. Then

Options

A. `4t = 6 + 0.6 t^(2)`
B. `1.2 t^(2) = 4t`
C. `4t^(2) = 1.2t`
D. `6 + 4t = 0.2t^(2)`

Answer

A

Explanation

Let us draw the figure for given situation,

`rArr 4t = 6 + (1)/(2) xx 1.2 xx t^(2)`
`rArr 4t = 6 + 0.6 t^(2)`