The position $x$ of a particle varies with time as $x = at^2 - bt^3$,where $a$ and $b$ are constants. The acceleration of the particle will be zero at:

$A$ point moves in a straight line so that its displacement $x \ m$ at time $t \ s$ is given by $x^2 = 1 + t^2$. Its acceleration in $m/s^2$ at a time $t \ s$ is

If $v = x^2 - 5x + 4$,find the acceleration of the particle when the velocity of the particle is zero.

Motion of a particle is given by the equation $s = (3t^3 + 7t^2 + 3t + 8) \ m$. The acceleration of the particle at $t = 1 \ s$ is: (in $m/s^2$)

An object with a mass $10 \, kg$ moves at a constant velocity of $10 \, m/s$. $A$ constant force then acts for $4 \, s$ on the object and gives it a speed of $2 \, m/s$ in the opposite direction. The acceleration produced in it is ........ $m/s^2$.

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