Figure shows the position-time graph of a particle of mass $4 \,kg$. What is the

$(a)$ force on the particle for $t\, <\, 0, t \,> \,4\; s, 0 \,<\, t \,< \,4\; s$?

$(b)$ impulse at $t=0$ and $t=4 \;s ?$ (Consider one-dimensional motion only).

$(a)$ For $t\,<\,0$

It can be observed from the given graph that the position of the particle is coincident with the time axis. It indicates that the displacement of the particle in this time interval is zero. Hence, the force acting on the particle is zero.

For $t \,>\, 4\; s$

It can be observed from the given graph that the position of the particle is parallel to the time axis. It indicates that the particle is at rest at a distance of $3 m$ from the origin. Hence, no force is acting on the particle.

For $0 \,<\, t \,<\, 4\;s$

$(b)$ At $t=0$ Impulse $=$ Change in momentum

$=m v-m u$

Mass of the particle, $m=4 \,kg$

Initial velocity of the particle, $u=0$

Final velocity of the particle, $v=\frac{3}{4} \,m / s$

$\therefore$ Impulse $=4\left(\frac{3}{4}-0\right)=3 \,kg\, m / s$

At $t=4\, s$

Initial velocity of the particle, $u=\frac{3}{4} \,m / s$

Final velocity of the particle, $v=0$

$\therefore$ Impulse $=4\left(0-\frac{3}{4}\right)=-3\, kg\, m / s$