$A$ 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 $...........\,s$.

If the velocity of a particle is given by $v = (180 - 16x)^{1/2} \text{ m/s}$,then its acceleration will be ....... $\text{m/s}^2$.

$A$ particle moving along the $x-$axis has acceleration $f$ at time $t$,given by $f = f_0(1 - t/T)$,where $f_0$ and $T$ are constants. The particle at $t = 0$ has zero velocity. In the time interval between $t = 0$ and the instant when $f = 0$,the particle's velocity $(v_x)$ is:

$A$ car is moving with velocity $V$. It stops after applying brakes at a distance of $20 \ m$. If the velocity of the car is doubled,then what distance will it cover after applying the brakes (in $m$)?

The velocity $(v)$ of a particle starting from rest increases linearly with time $(t)$ as $v = 4t$,where $v$ is in $m s^{-1}$ and $t$ is in seconds. The distance covered by the particle in the first $4$ seconds is (in $m$)

$A$ body starts from rest and acquires a velocity of $10 \ m \ s^{-1}$ in $2 \ s$. What is the acceleration of the body and the distance travelled?