$A$ body starts from rest with an acceleration of $8\,m/s^2$. What distance will it cover in the $5^{th}$ second?

$A$ particle starting from rest and moving with a uniform acceleration along a straight line covers distances $a$ and $b$ in successive intervals of $p$ and $q$ seconds. The acceleration of the particle 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

Displacement $(x)$ of a particle is related to time $(t)$ as: $x = at + bt^2 - ct^3$,where $a, b$,and $c$ are constants of the motion. The velocity of the particle when its acceleration is zero is given by:

$A$ lift of mass $M = 500 \, kg$ is descending with a speed of $2 \, ms^{-1}$. Its supporting cable begins to slip,thus allowing it to fall with a constant acceleration of $2 \, ms^{-2}$. The kinetic energy of the lift at the end of a fall through a distance of $6 \, m$ will be $........... \, kJ$.

Acceleration of a body whose displacement follows the equation $3 s = 9 t + 5 t^2$ is in $m s^{-2}$. (in $/3$)