To cross the river straight
$$\eqalign{ & {V_s}\sin \theta = {V_r} \cr & \therefore \sin \theta = \frac{{{v_r}}}{{{v_s}}} = \frac{2}{4} = \frac{1}{2} \cr & \therefore \theta = {30^ \circ } \cr} $$
Direction of swimmer with respect to flow
$$ = {90^ \circ } + {30^ \circ } = {120^ \circ }$$

$$\eqalign{ & {\text{Given that }}F\left( t \right) = {F_0}{e^{ - bt}} \Rightarrow m\frac{{dv}}{{dt}} = {F_0}{e^{ - bt}} \cr & \frac{{dv}}{{dt}} = \frac{{{F_0}}}{m}{e^{ - bt}} \Rightarrow \int\limits_0^v {dv} = \frac{{{F_0}}}{m}\int\limits_0^t {{e^{ - bt}}\,dt} \cr & v = \frac{{{F_0}}}{m}\left[ {\frac{{{e^{ - bt}}}}{{ - b}}} \right]_0^t = \frac{{{F_0}}}{{mb}}\left[ { - \left( {{e^{ - bt}} - {e^{ - 0}}} \right)} \right] \cr & \Rightarrow v = \frac{{{F_0}}}{{mb}}\left[ {1 - {e^{ - bt}}} \right] \cr} $$

The resultant vector $$ = \left( {3\hat i + 4\hat j + 5\hat k} \right) + \left( {5\hat i + 3\hat j + 4\hat k} \right)$$
$$\eqalign{ & = 8\hat i + 7\hat j + 9\hat k \cr & \cos \theta = \frac{8}{{\sqrt {{8^2} + {7^2} + {9^2}} }} = 0.574 \cr} $$

It is clear from the diagram that the shortest distance between the ship $$A$$ and $$B$$ is $$PQ.$$

Here, $$\sin {45^ \circ } = \frac{{PQ}}{{OQ}} \Rightarrow PQ = 100 \times \frac{1}{{\sqrt 2 }} = 50\sqrt 2 \,m$$
Also, $${v_{AB}} = \sqrt {v_A^2 + v_B^2} = \sqrt {{{10}^2} + {{10}^2}} = 10\sqrt 2 \,km/h$$
So, time taken for them to reach shortest path is
$$t = \frac{{PQ}}{{{v_{AB}}}} = \frac{{50\sqrt 2 }}{{10\sqrt 2 }} = 5\,h$$

Velocity is rate of change of position vector, i.e.,
$$v = \frac{{dr}}{{dt}}$$
where, $$r$$ is the position vector.
$$\eqalign{ & \therefore v = \frac{d}{{dt}}\left[ {\left( {a\cos \omega t} \right)\hat i + \left( {a\sin \omega t} \right)\hat j} \right] \cr & = \left( { - a\omega \sin \omega t} \right)\widehat i + \left( {a\omega \cos \omega t} \right)\widehat j \cr & = \omega \left[ {\left( { - a\sin \omega t} \right)\widehat i + \left( {a\cos \omega t} \right)\hat j} \right] \cr} $$
Slope of position vector $$ = \frac{{a\sin \omega t}}{{a\cos \omega t}} = \tan \omega t$$
and slope of velocity vector $$ = \frac{{ - a\cos \omega t}}{{a\sin \omega t}} = - \frac{1}{{\tan \omega t}}$$
$$\therefore $$ Velocity is perpendicular to the displacement.
Alternative
Velocity vector and position vector are perpendicular to each other if their scalar product is zero, i.e.
$$v \cdot r = 0$$
$$\eqalign{ & {\text{Now,}}\,v \cdot r = \left[ {\left( { - a\omega \sin \omega t} \right)\widehat i + \left( {a\cos \omega t} \right)\hat j} \right]\left[ {\left( {a\cos \omega t} \right)\widehat i + \left( {a\sin \omega t} \right)\hat j} \right] \cr & = - {a^2}\omega \sin \omega t\cos \omega t + {a^2}\omega \sin \omega t\cos \omega t = 0 \cr} $$
$$ \Rightarrow $$ Velocity vector is perpendicular to position vector.

Since, speed is constant throughout the motion, so, it is a uniform circular motion. Therefore, its radial acceleration is given by
$$\eqalign{ & {a_r} = r{\omega ^2} \cr & = r{\left( {\frac{{2\pi n}}{t}} \right)^2} = r \times \frac{{4{\pi ^2}{n^2}}}{{{t^2}}} \cr & = \frac{{1 \times 4 \times {\pi ^2} \times {{\left( {22} \right)}^2}}}{{{{\left( {44} \right)}^2}}} \cr & = {\pi ^2}\,m/{s^2} \cr} $$
This acceleration is directed along radius of circle.

given $$x = {V_0}t + \frac{1}{2}a{t^2}$$
$$\eqalign{ & \frac{{dx}}{{dt}} = {V_0} + \frac{1}{2} \times a \times 2t = {V_0} + at \cr & \Rightarrow V = {V_0} + at \cr} $$
so, option (C) represents the correct graph.

Distance covered by a particle is zero only when it is at rest. Therefore, its displacement must be zero.

Horizontal range $$R = \frac{{{u^2}\sin 2\theta }}{g}.$$
Range is same for angle of projection $$\theta $$ and $$\left( {{{90}^ \circ } - \theta } \right).$$

Comparing the given equation with the equation of trajectory of a projectile,
$$\eqalign{ & y = x\tan \theta - \frac{{g{x^2}}}{{2{u^2}{{\cos }^2}\theta }} \cr & {\text{we get,}}\,\,\tan \theta = \frac{1}{{\sqrt 3 }} \Rightarrow \theta = {30^ \circ } \cr & {\text{and}}\,\,2{u^2}{\cos ^2}\theta = 20 \Rightarrow {u^2} = \frac{{20}}{{2{{\cos }^2}\theta }} = \frac{{40}}{3} \cr & {\text{Now,}}\,\,{R_{\max }} = \frac{{{u^2}}}{{\;g}} = \frac{{40}}{{3 \times 10}} = \frac{4}{3}\;m \cr} $$