Fig. shows the position-time graph of a particle of mass 4 kg. What is the (i) force acting on the particle for `t lt 0, t lt 4 s, 0 lt t lt 4 s`? (ii) impulse at t = 0 and t = 4 s? Assume that the motion is one dimensional.

Explanation

(a) When t lt 0 . As this part is horizontal , thus it can be concluded that distance covered by the particle is zero and hence force on the particle is zero .
When 0 lt t lt 4s . As OA has a constant slope, hence in this interval , particle moves with constant velocity ( 3/4 = 0.75 `ms^(-1)` . Hence force on the particle is zero.
When t gt 4s. As this portion shows that particle always remains at a distance of 3m from the origin i.e. the particle is at rest. Hence force on the particle is zero.
(b) Impulse at t = 0 . Here u = 0 , v = 0.75 `ms^(-1)` , M = 4 kg
`therefore` Impulse = total change in momentum = Mv - Mu
= M(v-u) = 4(0.75 - 0) = 3 kg `ms^(-1)` .
Impulse at t = 4s. Here u = 0.75 `ms^(-1)` , v = 0
`therefore` Impulse` = M(v - u) = 4 ( 0 -0.75) = - 3 kg `ms^(-1)`.