The displacement of a particle moving with uniform acceleration in time $t$ is given by $S = 30t + 5t^2$. Its initial velocity is $.......$ (in $m \ s^{-1}$)

$A$ particle is projected with velocity $v_0$ along the $x$-axis. The deceleration of the particle is proportional to the square of the distance from the origin,i.e.,$a = -\alpha x^2$. The distance at which the particle stops is:

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

Two cars $A$ and $B$ initially at rest are moving in the same direction with accelerations $a_1$ and $a_2$ respectively. After a certain time $t$, they achieve velocities $v_1$ and $v_2$ respectively and are separated by a distance of $50 \ m$. If $(a_1 - a_2) = 4 \ m \ s^{-2}$, then the quantity $(v_1 - v_2)$ will be: (in $m \ s^{-1}$)

$A$ car moving at a speed of $30 \, km/hr$ comes to a stop after traveling $8 \, m$ when the brakes are applied. If the same car is moving at a speed of $60 \, km/hr$,what distance (in $m$) will it travel before coming to a stop after the brakes are applied?

$A$ 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 $O$?