$A$ 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 varies with time as:

$A$ 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 $.... \ m$.

$A$ 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?

The distance $x$ covered in time $t$ by a body having initial velocity $u$ and having constant acceleration $a$ is given by $x=ut+\frac{1}{2}at^2$. This result follows from

An automobile travelling with a speed of $60\,km/h$ can brake to stop within a distance of $20\,m$. If the car is going twice as fast,i.e.,$120\,km/h$,the stopping distance will be ........... $m$.

The initial velocity of a particle is $u$ (at $t = 0$) and the acceleration is given by $a(t) = at$. Which of the following relations is valid?