$$\eqalign{ & \left| {{\text{Average velocity}}} \right| = \frac{{\left| {{\text{Displacement}}} \right|}}{{{\text{Time}}}} \cr & = \frac{{2r}}{t} \cr & = 2 \times \frac{1}{1} \cr & = 2\,m/s \cr} $$

$$\eqalign{ & {\text{Average}}\,{\text{acceleration}} = \frac{{{\text{change}}\,{\text{in}}\,{\text{velocity}}}}{{{\text{time interval}}}} \cr & = \frac{{\Delta \overrightarrow v }}{t} \cr & \overrightarrow {{v_1}} = 5\hat i,\overrightarrow {{v_2}} = 5\hat j \cr & \Delta \overrightarrow v = \left( {\overrightarrow {{v_2}} - \overrightarrow {{v_1}} } \right) = \sqrt {v_1^2 + v_2^2 + 2{v_1}{v_2}\cos {{90}^ \circ }} \cr & = \sqrt {{5^2} + {5^2} + 0} \cr & \left[ {{\text{As}}\,\,\left| {{v_1}} \right| = \left| {{v_2}} \right| = 5\,m/s} \right] = 5\sqrt 2 \,m/s \cr & {\text{Avg}}{\text{.}}\,{\text{acc}}{\text{.}}\, = \frac{{\Delta \overrightarrow v }}{t} = \frac{{5\sqrt 2 }}{{10}} = \frac{1}{{\sqrt 2 }}m/{s^2} \cr & \Rightarrow \tan \theta = \frac{5}{{ - 5}} = - 1 \cr} $$
which means $$\theta $$ is in the second quadrant. (towards north-west)

Distance travelled in the $$n$$^th second is given by
$$\eqalign{ & {t_n} = u + \frac{a}{2}\left( {2n - 1} \right) \cr & {\text{put}}\,u = 0,a = \frac{4}{3}m{s^{ - 2}},n = 3 \cr & \therefore d = 0 + \frac{4}{{3 \times 2}}\left( {2 \times 3 - 1} \right) = \frac{4}{6} \times 5 = \frac{{10}}{3}m \cr} $$

$$\eqalign{ & {\text{Time}} = \frac{{{\text{total}}\,{\text{length}}}}{{{\text{relative velocity}}}} \cr & = \frac{{X + X}}{{20 + 20}} = \frac{{2X}}{{40}}s \cr & = 2s \cr} $$

We should know the displacement in this time.
$$\eqalign{ & \overrightarrow S = \overrightarrow u t + \frac{{\overrightarrow a {t^2}}}{2}\,\left( {{\text{we take upward as positive}}} \right) \cr & S = 30 \times 4 - 10 \times 4 \times \frac{4}{2} = 40\,m. \cr} $$
The average velocity will be $$10\,m/\sec .$$

Given, the position $$x$$ of a particle w.r.t. time $$t$$ along $$x$$-axis
$$x = 9{t^2} - {t^3}\,......\left( {\text{i}} \right)$$
Differentiating Eq. (i), w.r.t. time, we get speed, i.e.
$$\eqalign{ & v = \frac{{dx}}{{dt}} = \frac{d}{{dt}}\left( {9{t^2} - {t^3}} \right) \cr & {\text{or}}\,v = 18t - 3{t^2}\,......\left( {{\text{ii}}} \right) \cr} $$
Again differentiating Eq. (ii), with respect to time, we get acceleration, i.e.
$$\eqalign{ & a = \frac{{dv}}{{dt}} = \frac{d}{{dt}}\left( {18t - 3{t^2}} \right) \cr & {\text{or,}}\,a = 18 - 6t\,......\left( {{\text{iii}}} \right) \cr} $$
Now, when speed of particle is maximum, its acceleration is zero, i.e.
$$\eqalign{ & a = 0 \cr & {\text{i}}{\text{.e}}{\text{.}}\,18 - 6t = 0 \cr & {\text{or}}\,t = 3\,s \cr} $$
Putting in Eq. (i), we obtain position of particle at the time
$$\eqalign{ & x = 9{\left( 3 \right)^2} - {\left( 3 \right)^3} \cr & = 9\left( 9 \right) - 27 \cr & = 81 - 27 \cr & = 54\,m \cr} $$

Time taken by the stone to reach the water level $${t_1} = \sqrt {\frac{{2h}}{g}} $$
Time taken by sound to come to the mouth of the well, $${t_2} = \frac{h}{v}$$
$$\therefore $$ Total time $${t_1} + {t_2} = \sqrt {\frac{{2h}}{g}} + \frac{h}{v}$$

Horizontal components of velocity at $$O$$ and $$P$$ are equal.

$$\eqalign{ & \therefore v\cos \left( {{{90}^ \circ } - \theta } \right) = u\cos \theta \cr & {\text{or}}\,\,v\sin \theta = u\cos \theta \cr & {\text{or}}\,v = u\cot \theta \cr & {\text{At}}\,\,P,\frac{{v_T^2}}{R} = {a_c}; \cr & \frac{{{u^2}{{\cot }^2}\theta }}{{g\sin \theta }} = R \cr} $$

According to pythagorus theorem
$${V_{rm}} = \sqrt {V_r^2 + V_m^2} $$

As slope is same, so velocity of $$A$$ and $$B$$ will be equal