$A$ particle is moving in a straight line and passes through a point $O$ with a velocity of $6\;m/s$. The particle moves with a constant retardation of $2\;m/s^2$ for $4\;s$ and thereafter moves with constant velocity. How long after leaving $O$ does the particle return to $O$?

$A$ 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 .............. $m/s$.

Object $A$ starts from rest with a constant acceleration $a$. Object $B$ starts from the same position and moves in the same direction as $A$ with a constant velocity $v$. If both meet after time $t$,then $t =$

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 $\text{ m/s}^2$)

$A$ body starts from rest with uniform acceleration $a$. Its velocity after $n \ s$ is $v$. The displacement of the body in the last $3 \ s$ is: (Assume total time of journey is $n \ s$)

The stopping distance of a vehicle moving with a given acceleration is proportional to the ....... of its speed.