$A$ particle is moving along a line according to the law $S = t^3 - 3t^2 + 4t - 2$,where $S$ is measured in meters and $t$ is measured in seconds. Then the velocity (in $m/s$) of the particle when its acceleration is zero is:

If a particle moves in a straight line according to the law $x = a \sin (\sqrt{\lambda} t + b)$, then the particle will come to rest at two points whose distance is [symbols have their usual meaning]

The distance $s$ in meters travelled by a particle in $t$ seconds is given by $s = \frac{2 t^3}{3} - 18 t + \frac{5}{3}$. The acceleration when the particle comes to rest is (in $m/s^2$)

$A$ street light is at the top of a $12 \ m$ pole. $A$ man $2 \ m$ tall walks away from the pole towards a wall $12 \ m$ away from the pole at a speed of $1/2 \ m/s$. The rate at which his shadow on the wall is decreasing when he is $8 \ m$ from the wall is:

If the path of a moving point is the curve $x = at$,$y = b \sin(at)$,then its acceleration at any instant

$A$ particle is moving in a straight line according to the equation $s = 45t + 11t^2 - t^3$. The time at which it will come to rest is ......... $sec$.