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 $m$)

$A$ particle moves in a straight line and its position $x$ at time $t$ is given by $x^2 = 2 + t$. Its acceleration is given by

$A$ car moving at a speed $V$ is stopped in a certain distance $s$ when the brakes produce a deceleration $a$. If the speed of the car is $nV$,what must be the deceleration of the car to stop it in the same distance $s$ and in the same time $t$?

$A$ body moves on a frictionless plane starting from rest. If $S_{n}$ is the distance moved between $t=n-1$ and $t=n$,and $S_{n-1}$ is the distance moved between $t=n-2$ and $t=n-1$,then the ratio $\frac{S_{n-1}}{S_n}$ is $\left(1-\frac{2}{x}\right)$ for $n=10$. The value of $x$ is:

$A$ particle moves along a straight line such that its displacement at any time $t$ is given by $s = (t^3 - 6t^2 + 3t + 4) \ m$. Find the velocity of the particle in $m/s$ when its acceleration is zero.

$A$ particle is moving in a straight line. The variation of position $x$ as a function of time $t$ is given as $x = (t^3 - 6t^2 + 20t + 15) \ m$. The velocity of the body when its acceleration becomes zero is ........... $m/s$.