Draw the $x-t$ graph for negative acceleration.

If the velocity of a particle moving along a straight line with uniform acceleration is given by $V = \sqrt{196 - 16x} \text{ m/s}$, then its acceleration is ($x$ is the displacement of the particle). (in $\text{ m/s}^2$)

If the velocity of a particle is $v = At + Bt^2$,where $A$ and $B$ are constants,then the distance travelled by it between $1 \ s$ and $2 \ s$ is

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

$m$ 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 $x = 0$.

An object moves in a straight line with deceleration whose magnitude varies with velocity as $3 v^{2/3}$. If at an initial point, the velocity is $8 \,m/s$, then the distance travelled by the object before it stops is (in $\,m$)