Two men $P$ and $Q$ are standing at corners $A$ and $B$ of square $ABCD$ of side $8 \mathrm{m}.$ They start moving along the track with constant speed $2 \mathrm{m} {\mathrm{s}}^{-1}$ and $10 \mathrm{m} {\mathrm{s}}^{-1}$ respectively. Find the time when they will meet for the first time.

A man in a balloon throws a stone downwards with a speed of $5 \mathrm{m} {\mathrm{s}}^{-1}$ with respect to the balloon. The balloon is moving upwards with a constant acceleration of $5 \mathrm{m} {\mathrm{s}}^{-2}.$ Then velocity of the stone relative to the man after $2$ second is:

Two particles are moving along a straight line as shown. The velocity of approach between $A$ and $B$ is

Car $A$ and car $B$ move on a straight road and their velocity versus time graphs are as shown in the figure. Comparing the motion of car $A$ in between $t=0$ to $t=8 \mathrm{s}$ and motion of car $B$ in between  $t=0$ to $t=7$ $\mathrm{s}$, pick up the correct statement.