$A$ car moving along a straight highway with a speed of $126 \; km/h$ is brought to a stop within a distance of $200 \; m$. What is the retardation of the car (assumed uniform in $m/s^2$)?

$A$ car moving at a speed of $50 \, km/hr$ stops after covering a distance of $6 \, m$ upon applying brakes. If the same car is moving at a speed of $100 \, km/hr$,what distance (in $m$) will it cover after applying the brakes?

The motion of a particle along a straight line is described by the equation $x = 8 + 12t - t^3$,where $x$ is in meters and $t$ is in seconds. The retardation of the particle when its velocity becomes zero is .......... $m/s^2$.

Two cars leave one after the other and travel with an acceleration of $0.4\, m/s^2$. Two minutes after the departure of the first,the distance between the cars becomes $1.9\, km$. The time interval between the departures of the cars is ........$s$.

The velocity of a particle is $v = v_{0} + gt + Ft^{2}$. Its position is $x = 0$ at $t = 0$; then its displacement after time $t = 1$ is:

$A$ body starts with an initial velocity of $10\,m/s$ and moves along a straight line with constant acceleration. When the velocity of the particle is $50\,m/s$,the acceleration is reversed in direction. The velocity of the particle when it again reaches the starting point is $............\,m/s$.