$A$ particle starts moving rectilinearly at time $t = 0$ such that its velocity $v$ changes with time $t$ according to the equation $v = t^2 - t$,where $t$ is in seconds and $v$ is in $m/s$. The time interval for which the particle retards is

Stopping distance of vehicles: When brakes are applied to a moving vehicle,the distance it travels before stopping is called stopping distance. It is an important factor for road safety and depends on the initial velocity $(v_0)$ and the braking capacity,or deceleration,$-a$ that is caused by the braking. Derive an expression for stopping distance of a vehicle in terms of $v_0$ and $a$.

$A$ bus moving along a straight highway with a speed of $72 \,km/h$ is brought to a halt within $4 \,s$ after applying the brakes. The distance travelled by the bus during this time (assume the retardation is uniform) is . . . . . . $m$.

Which of the following decreases,in motion on a straight line,with constant retardation?

$A$ particle is moving in a straight line and passes through a point $O$ with a velocity of $6\;m/s$. The particle moves with a constant retardation of $2\;m/s^2$ for $4\;s$ and thereafter moves with constant velocity. How long after leaving $O$ does the particle return to $O$?

Particle $A$ is moving along the $x$-axis. At time $t=0$,it has a velocity of $10\,m/s$ and an acceleration of $-4\,m/s^2$. Particle $B$ has a velocity of $20\,m/s$ and an acceleration of $-2\,m/s^2$. Initially,both particles are at the origin. At time $t=2\,s$,the distance between the two particles is $.............\,m$.