Question

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

Answer 1

(a) Let us draw the figure for given situation,

`implies4t=6+1/2xx1.2xxt^(2)`
`implies 4t=6+0.6t^(2)`