Figure shows the position - time graph of a particle of mass 4 kg, then
(a) force on the particle for `r lt 0, t gt 4 s, 0 lt t lt 4 s` ?
(b) impulse at t = 0 and t = 4 s. (consider one - dimensional motion only)

Explanation

(a) (i) For `t lt 0`, in the position time graph is at O which means displacement of the particle is zero. i.e., particle is at rest at the origin. Hence force on the particle must be zero.
(ii) For `0 lt lt 4` s, the position time graph OA has a constant slope. Therefore velocity of the particle is constant in this interval i.e., particle has zero acceleration. Hence force on he particle must be zero.
(iii) Fro `t gt 4` s, the position time graph AB is parallel to time axis. Therefore, the particle remains at a distance 3 m from the origin, i.e., it is at rest. Hence force on the particle is zero.
(b) (i) impulse at t = 0
We know, Impulse = change in linear momentum,
Before t = 0 particle is at rest i.e., u = 0.
After t = 0, particle has a constant velocity
`v=(3)/(4)=0.75 m//s`
`therefore` Impulse `= m(v-u)=m(0.75-0)=3kgm//s`
(ii) `therefore` Impulse at t = 4 s.
Before t = 4s,
particle has a constant velocity u = 0.75 m/s
After t = 4s, particle is at rest i.e., v = 0
Impulse `= m(v-u)=4(0-0.75)=-3 kg m//s`.