A 100 m long train crosses a man travelling at `5 kmh^(-1)` , in opposite direction, in `7.2 s` then the velocity of train is

Options

A. In the interval `3 lt t lt 5`, the particle is moving in +x direction
B. The particle reverses its direction of motion twice in entire motion if it starts at t=0
C. The average acceleration from `1 le t le 2` second is `6 m//s^(2)`
D. In the interval `5 le t le 6` seconds, the distance travelled is equal to the displacement.

Answer

A, D

Explanation

`x=t^(3)-3t^(2)-9t+5. x(5) gt 0` and `x(3) gt 0`
so `[A] v=dx//dt=3t^(2)-6t-9`
`rArr t=-1, 3` so `t=3`
Hence particle reverses its direction only once average acc.= change in velocity `//`time.
In interval `(t=3 " to " t=6)`, particle does not reverse its velocity and also moves in a straight line so distance =displacement.