$A$ 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 .......... $m/s^2$.

$A$ particle is projected with velocity $v_0$ along the $x-$axis,and its deceleration is given by $a = -\alpha x^2$. The distance at which the particle stops is

An object moving along the $x$-axis with a uniform acceleration has a velocity $\vec{v} = (12 \ cm \ s^{-1}) \hat{i}$ at $x = 3 \ cm$. After $2 \ s$,if it is at $x = -5 \ cm$,then its acceleration is:

$A$ 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 $m/s^2$)

The velocity of a car accelerating uniformly increases from $20 \ m/s$ to $80 \ m/s$ while covering a distance of $200 \ m$. The time taken is $.... \ s$.

Velocity of a body moving along a straight line with uniform acceleration $a$ reduces by $\frac{3}{4}$ of its initial velocity in time $t_0$. The total time of motion of the body till its velocity becomes zero is