$A$ particle moves according to the equation $x = at^2 - bt^3$. At what time will its acceleration be zero?

$A$ small block slides down on a smooth inclined plane,starting from rest at time $t=0$. Let $S_{n}$ be the distance travelled by the block in the interval $t=n-1$ to $t=n$. Then,the ratio $\frac{S_{n}}{S_{n+1}}$ is

$A$ particle moves according to the equation $y = a + bt + ct^2 - dt^4$. What are the initial velocity and initial acceleration of the particle?

If the displacement of a particle varies with time as $\sqrt{x} = t + 7$,then

$A$ body $A$ moves with a uniform acceleration $a$ and zero initial velocity. Another body $B$ starts from the same point and moves in the same direction with a constant velocity $v$. The two bodies meet after a time $t$. The value of $t$ is:

$A$ particle starts from the origin at time $t=0$ and moves in the positive $x$-direction. Its velocity $v$ varies with time as $v=10t \text{ cm/s}$. The distance covered by the particle in $8 \text{ s}$ will be: (in $\text{ cm}$)