$A$ graph between the square of the velocity of a particle and the distance $s$ moved by the particle is shown in the figure. The acceleration of the particle is $...........m/s^2$.

The displacement of a particle is proportional to the cube of time elapsed. How does the acceleration of the particle depend on time?

The ratio of displacement in $n$ seconds and in the $n^{th}$ second for a particle moving in a straight line under constant acceleration starting from rest is:

The displacement of a particle moving in a straight line is given by the expression $x = A t^3 + B t^2 + C t + D$ in metres, where $t$ is in seconds and $A, B, C$ and $D$ are constants. The ratio between the initial acceleration and initial velocity is

$m$ mass particle is constrained to move on the $x$-axis. $A$ force $F$ acts on the particle,always pointing toward the equilibrium position $E$. The magnitude of $F$ is constant except at $E$ where it is zero. The particle is displaced a distance $A$ to the left of $E$ and released from rest at $t = 0$. Find the minimum time taken to reach from $x = -A/2$ to $x = 0$.

Stopping distance of vehicles: When brakes are applied to a moving vehicle,the distance it travels before stopping is called stopping distance. It is an important factor for road safety and depends on the initial velocity $(v_0)$ and the braking capacity,or deceleration,$-a$ that is caused by the braking. Derive an expression for stopping distance of a vehicle in terms of $v_0$ and $a$.