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` ?
.

Options

A. `(2)/(3)"kgms"^(-1)`
B. `-(2)/(3)"kgms"^(-1)`
C. `(3)/(2)"kgm"s^(-1)`
D. `-(3)/(2)"kgm"s^(-1)`

Answer

D

Explanation

At t=4s, the body has constant velocity
`u=(3)/(4)ms^(-1)`
After t=4s, the body is at rest i.e. v=0
`therefore "Impulse" =m(v-u)`
`=2kg(0-(3)/(4)ms^(-1))=-(3)/(2) "kg ms"^(-1)`