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

Two cars leave one after the other and travel with an acceleration of $0.4\, m/s^2$. Two minutes after the departure of the first,the distance between the cars becomes $1.9\, km$. The time interval between the departures of the cars is ........$s$.

$A$ particle moves for $8 \, s$. It first accelerates from rest and then retards to rest. If the retardation is $3 \, \times$ the acceleration, then the time for which it accelerates is: (in $s$)

Stopping distance of vehicles: When brakes are applied to a moving vehicle,the distance it travels before stopping is called stopping distance. It is an important factor for road safety and depends on the initial velocity $(v_0)$ and the braking capacity,or deceleration,$-a$ that is caused by the braking. Derive an expression for stopping distance of a vehicle in terms of $v_0$ and $a$.

$A$ particle moves with constant acceleration. Let $v_1, v_2, v_3$ be the average velocities in successive time intervals $t_1, t_2$ and $t_3$. Then:

$A$ particle moves a distance $x$ in time $t$ according to the equation $x = (t + 5)^{-1}$. The acceleration of the particle is proportional to: