A lift starts from the top of a mine shaft and descends with a constant speed of $10 \mathrm{m} {\mathrm{s}}^{-1}$. $4 \mathrm{s}$ later a boy throws a stone vertically upwards from the top of the shaft with a speed of $30 \mathrm{m} {\mathrm{s}}^{-1}$. If stone hits the lift at a distance $x$ below the shaft write the value of $\frac{x}{3}\left(\mathrm{m}\right) \left[g=10 \mathrm{m} {\mathrm{s}}^{-2}\right.$,
Give value of $20\sqrt{6}=49)$

The acceleration time plot for a particle (starting from rest) moving on a straight line is shown in the figure. For given time interval,

A particle moves along the $Y$ -axis and its $y$ -coordinate $\left(y\right)$changes with time $\left(t\right)$ as $y=u(t-2)+a(t-2{)}^2$.

The graph below gives the coordinator of a particle travelling along the $X-$axis as a function of $\tan \theta$. $AM$ is the tangent to the curve at the starting moment and $BN$ is tangent at the end moment $\left({\theta }_1={\theta }_2=120°\right)$

The direction $(i$ or $-i)$ of acceleration during the first $10$ seconds is

Three elephants $A$, $B$ and $C$ are moving along a straight line with constant speed in the same direction, as shown in the figure. Speed of $A$ is $5 \mathrm{m} {\mathrm{s}}^{-1}$ and speed of $C$ is $10 \mathrm{m} {\mathrm{s}}^{-1}$. Initially, the separation between $A$ and $B$ is $d$ and between $B$ and $C$ is also $d$. When $B$ catches $C$ separation between $A$ and $C$ becomes $3d$. Find the speed of $B \left(\mathrm{in} \mathrm{m} {\mathrm{s}}^{-1}\right)$