$A$ car moving with a speed of $40 \, km/h$ can be stopped by applying brakes after at least $2 \, m$. If the same car is moving with a speed of $80 \, km/h$,what is the minimum stopping distance in meters?

$A$ car moving with a velocity $6.25 \,ms^{-1}$ is decelerated with $2.5 \sqrt{v} \,ms^{-2}$ (where $v$ is the instantaneous velocity). The time taken by the car to come to rest is: (in $\,s$)

The displacement of a particle is proportional to the cube of time elapsed. How does the acceleration of the particle depend on time?

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?

$A$ car moving with a velocity of $20 \,m \,s^{-1}$ is stopped in a distance of $40 \,m$. If the same car is travelling at double the velocity, the distance travelled by it for the same retardation is (in $\,m$)

$A$ body starts from the origin and moves along the $X$-axis such that the velocity at any instant is given by $v = (4t^3 - 2t)$,where $t$ is in $s$ and velocity is in $m/s$. What is the acceleration of the particle when it is $2\, m$ from the origin? (in $m/s^2$)