$A$ particle is moving with constant acceleration $a$. The following graph shows the $v^{2}$ versus $x$ (displacement) plot. The acceleration of the particle is $...... \text{m/s}^{2}$.

The displacement and the increase in the velocity of a moving particle in the time interval of $t$ to $(t+1) s$ are $125 \ m$ and $50 \ m/s$,respectively. The distance travelled by the particle in $(t+2)^{th} s$ is . . . . . . $m$.

The velocity of a particle moving in the positive direction of the $x$-axis varies as $v = 5 \sqrt{x}$. Assuming that at $t = 0$,the particle was at $x = 0$,what is the acceleration of the particle in $m/s^2$?

$A$ bullet fired into a target loses one-third of its velocity after travelling a distance $x$ meters into the target. If the bullet comes to rest by travelling a further distance $x^{\prime}$, then the ratio $\frac{x^{\prime}}{x}$ is

$A$ particle initially at rest moves along the $x$-axis. Its acceleration varies with time as $a = 4t$. If it starts from the origin,the distance covered by it in $3 \ s$ is $........... \ m$.

$A$ particle moves according to the equation $x = at^2 - bt^3$. At what time will its acceleration be zero?