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

$A$ particle initially at rest starts moving from reference point $x=0$ along the $x$-axis,with velocity $v$ that varies as $v=4 \sqrt{x} \ m/s$. The acceleration of the particle is . . . . . . $m/s^2$.

On what factors does the stopping distance depend?

$A$ particle starts from rest at $x=0 \, m$ with an acceleration of $1 \, m/s^2$. At $t = 5 \, s$,it receives an additional acceleration in the same direction as its motion. At $t = 10 \, s$,its speed and position are $v$ and $x$,respectively. Had the additional acceleration not been provided,its speed and position would have been $v_0$ and $x_0$,respectively. It is found that $x - x_0 = 12.5 \, m$. Then one can conclude that $v - v_0$ is .............. $m/s$.

The relation between time $t$ and distance $x$ for a moving body is given as $t = m x^{2} + n x$,where $m$ and $n$ are constants. The retardation of the motion is:

The acceleration of a particle is increasing linearly with time $t$ as $bt$. The particle starts from the origin with an initial velocity ${v_0}$. The distance travelled by the particle in time $t$ will be