The figure shows the position-time graph of a particle of mass $4 \,kg$. What is the
$(a)$ force on the particle for $t < 0$,$t > 4 \,s$,and $0 < t < 4 \,s$?
$(b)$ impulse at $t = 0$ and $t = 4 \,s$? (Consider one-dimensional motion only).

(N/A) For $t < 0$:
The graph shows that the position of the particle is $x = 0$ (coincident with the time axis). This indicates that the particle is at rest. Hence,the velocity is zero,and the force acting on the particle is $F = 0$.
For $t > 4 \,s$:
The graph shows that the position of the particle is constant at $x = 3 \,m$ (parallel to the time axis). This indicates that the particle is at rest. Hence,the velocity is zero,and the force acting on the particle is $F = 0$.
For $0 < t < 4 \,s$:
The graph is a straight line with a constant slope. The velocity $v = \frac{\Delta x}{\Delta t} = \frac{3 - 0}{4 - 0} = 0.75 \,m/s$. Since the velocity is constant,the acceleration is zero. Hence,the force acting on the particle is $F = m \cdot a = 0$.
$(b)$ Impulse is defined as the change in momentum,$J = \Delta p = m(v - u)$.
At $t = 0 \,s$:
Initial velocity $u = 0$,final velocity $v = 0.75 \,m/s$.
$J = 4 \,kg \times (0.75 - 0) \,m/s = 3 \,kg \cdot m/s$.
At $t = 4 \,s$:
Initial velocity $u = 0.75 \,m/s$,final velocity $v = 0$.
$J = 4 \,kg \times (0 - 0.75) \,m/s = -3 \,kg \cdot m/s$.