$A$ particle moving with uniform acceleration travels $24\, m$ and $64\, m$ in the first two consecutive intervals of $4\, s$ each. Its initial velocity is ....... $m/s$.

$A$ moving body is covering distances which are proportional to the square of the time. Then the acceleration of the body is

$A$ bullet is fired into a fixed target and loses one-third of its velocity after traveling $4 \ cm$. It penetrates a further $D \times 10^{-3} \ m$ before coming to rest. The value of $D$ is:

$A$ particle starts moving from rest in a straight line with constant acceleration. After time $t_0$,the acceleration changes its sign (becomes exactly opposite to the initial direction),while remaining the same in magnitude. Determine the time from the beginning of the motion at which the particle returns to its initial position.

$A$ particle returns to the starting point after $10\,s$. If the rate of change of velocity during the motion is constant,then its location after $7\,s$ will be same as that after $...........\,s$.

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