A ball projected vertically upwards with a velocity ' $V$ ' passes through a point $P$ in its upward journey in a time of ' $x$ ' seconds. From there, the time in which the ball again passes through the same point $P$ is

Two smooth inclined planes $A$ and $B$ each of height 20 m have angles of inclination $30^{\circ}$ and $60^{\circ}$ respectively. If $t_1$ and $t_2$ are respectively the times taken by two blocks to reach the bottom of the planes $A$ and $B$ from the top, then $t_1-t_2=$ (Acceleration due to gravity $=10 \mathrm{~ms}^{-2}$ )

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

The displacement $(x)$ and time $(t)$ graph of a particle moving along a straight line is shown in the figure. The average velocity of the particle in the time of 10 s is