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 $m$)

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 ........... $m$.

In a car race on a straight road,car $A$ takes a time $t$ less than car $B$ at the finish and passes the finishing point with a speed $v$ more than that of car $B$. Both the cars start from rest and travel with constant acceleration $a_1$ and $a_2$ respectively. Then $v$ is equal to

The acceleration of a particle is increasing linearly with time $t$ as $bt$. The particle starts from the origin with an initial velocity ${v_0}$. The distance travelled by the particle in time $t$ will be

The position of a particle as a function of time $t$ is given by $x(t) = at + bt^2 - ct^3$,where $a, b,$ and $c$ are constants. When the particle attains zero acceleration,its velocity will be:

For a train engine moving with a speed of $20 \; m/s$,the driver must apply brakes at a distance of $500 \; m$ before the station for the train to come to rest at the station. If the brakes were applied at half of this distance,the train engine would cross the station with a speed of $\sqrt{x} \; m/s$. The value of $x$ is $..............$ (Assuming the same retardation is produced by the brakes).