Shows position time graph of a particle of mass `100g` Find the impulse at `t = 0` and at `t = 4 s`
.

Answer

`0.05 kg m//s; - 0.05 kg m//s` .

Explanation

Shows that at `t=0` particle is at rest
after `t=0` the particle has unifrom velocity.
`upsilon =(dx)/(dt) =(2)/(4) =0.5m//s`
`:.` Impulse at t = 0 is
`=m (upsilon -u)`
`=0.1 (0.5 -0)`
`=0.5 kg ms^(-1)`
Again Before `t =4 s` particle is moving with a unifrom velocity `u= 0.5m//s` and after `t = 4s`
particle comes to rest i.e `upsilon =0`
Impulse at `t=4s`
`=m(upsilon - u) = 0.1 (0 -0.5)`
`= -0.05 kg ms^(-1)` .