The position time graph of a body of mass `2 kg` is as given in What is the impulse on the body at ` t = 0 ` s and `t = 4 s` ?
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Explanation

Here, ` m = 2 kg `
As is clear from position time graph in
At ` t = 0 , x = 0 ` body is at rest . Therefore impulse of body (at t=0)=0
From ` t = 0 ` to `t = 4 ` s position time graph of body is a st line with positive slope. The body moves with a unifrom velocity.
Beyoud ` t = 4 ` s position time graph is parallel to time axis Therefore, body comes to rest
Now initial vel of body ` (u) = tan theta = (3)/(4) m//s` final vel of body `(upsilon) = 0 `
`:.` Impulse (at t = 4 s) = Change in momentum ` = m (upsilon - u ) = 2 (0-(3)/(4)) = - (3)/(2) kg ms^(-1)` .