$A$ car accelerates from rest at a constant rate of $2 \ m/s^2$ for some time. Then it retards at a constant rate of $4 \ m/s^2$ and comes to rest. If it remains in motion for $3 \ s$,then the total displacement covered by it is $.... \ m$.

If the velocity of a body related to displacement $x$ is given by $v = \sqrt{5000 + 24x} \; \text{m/s}$,then the acceleration of the body is $\dots \dots \; \text{m/s}^2$.

$A$ particle located at $x = 0$ at time $t = 0$ starts moving along the positive $X$-direction with a velocity $v$ that varies as $v = \alpha \sqrt{x}$. The displacement of the particle with time is proportional to

$A$ motor car moving with a uniform speed of $20\,m/s$ comes to a stop on the application of brakes after travelling a distance of $10\,m$. Its acceleration is .......... $m/s^2$.

$A$ car starts at time $t=0$ from an initial speed of $10 \,m/s$ and accelerates uniformly with $2 \,m/s^2$ on a straight road for time $0 \leq t \leq 10 \,s$. Let $S_1$ and $S_2$ be the distance covered by the car in time $3 \leq t \leq 4 \,s$ and $4 \leq t \leq 5 \,s$ respectively. The ratio $\frac{S_2}{S_1}$ is

$A$ 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 $........... \, kJ$.