The position $x$ of a particle varies with time $t$ as $x = at^2 - bt^3$. The acceleration of the particle will be zero at time $t$ equal to

Which physical quantity can be found by first differentiation and second differentiation of position vector?

$A$ body is moving with variable acceleration $a$ along a straight line. The average acceleration of the body in the time interval $t_1$ to $t_2$ is

$A$ particle moves along the $x$-axis and its displacement $x$ varies with time $t$ according to the equation $x = c_0(t^2 - 2) + c(t - 2)^2$,where $c_0$ and $c$ are constants of appropriate dimensions. Which of the following statements is correct?

If the displacement ($s$ in metre) of a moving particle in terms of time ($t$ in second) is $s = t^3 - 6t^2 + 18t + 9$,then the minimum velocity attained by the particle is (in $m \ s^{-1}$)

The displacement of a particle,moving in a straight line,is given by $s = 2t^2 + 2t + 4$,where $s$ is in metres and $t$ is in seconds. The acceleration of the particle is ........ $m/s^2$.