Two particles $P$ and $Q$ are initially $40 \mathrm{m}$ apart $P$ behind $Q.$ Particle $P$ starts moving with a uniform velocity $10 \mathrm{m} {\mathrm{s}}^{-1}$ towards $Q.$ Particle $Q$ starting from rest has an acceleration $2 \mathrm{m} {\mathrm{s}}^{-2}$ in the direction of velocity of $P.$Then the minimum distance between $P$ and $Q$ will be $n \mathrm{m}.$ The value of $n$ is

A ball is projected vertically upwards at$t=0$. The ball is at same height at time $4 \mathrm{s}$ and $8 \mathrm{s}$. The maximum height attained by the ball is 

$\left(g=10 \mathrm{m} {{\mathrm{s}}^{-}}^2\right)$

The velocity-position graph of a particle moving in a straight line along$x$-axis is given below. Acceleration of particle at$x=2 \mathrm{m}$ is

The acceleration-position graph of a particle moving in a straight line along $x$-axis is shown in figure. The particle starts from rest at origin. The maximum speed of particle is

The acceleration-time graph of a particle moving in a straight line along $x$-axis is given below. If initial velocity of the particle is $4 \mathrm{m} {\mathrm{s}}^{-1}$, then velocity at $t=2 \mathrm{s}$ is