Question

If ${\overrightarrow{v}}_{wm}$ is the velocity of a man relative to water, ${\overrightarrow{v}}_w$ is the velocity of water, and ${\overrightarrow{v}}_m$ is the velocity of man relative to ground, match the following:

	Column I		Column II
I.	Minimum distance for $v_{mw}>v_w$	a.	$\theta =si{n}^{-1}\left(\frac{v_{mw}}{v_w}\right)$
II.	Minimum time for $v_{mw}\ge v_w$	b.	${\overrightarrow{v}}_m\perp {\overrightarrow{v}}_w$
III.	Minimum distance for $v_{mw}<v_w$	c.	${\overrightarrow{v}}_{mw}\perp {\overrightarrow{v}}_w$
iv.	Minimum time for $v_{mw}<v_w$	d.	$\theta =si{n}^{-1}\frac{v_w}{v_{mw}}$

where $\theta$ is the angle between ${\overrightarrow{v}}_{mw}$ and the width of the river.

Answer 1

$i.\to b,d.;ii.\to c.;iii.\to a.;iv.\to c$.
Distance will be minimum because man will reach from $A$ to $B$ directly.
$\sin \theta =\frac{v_{\omega }}{v_{m\omega }}$ and ${\overrightarrow{v}}_m\perp {\overrightarrow{v}}_{\omega }$
Hence $i\to b,d$
Time taken is minimum if ${\overrightarrow{v}}_{m\omega }$ is perpendicular to ${\overrightarrow{v}}_{\omega }$.
Hence $ii.\to c.,iv\to c$.
If ${\overrightarrow{v}}_{w\omega }<{\overrightarrow{v}}_{\omega }$, then drift or distance is shortest if
$\sin \theta =\frac{v_{m\omega }}{v\omega }$. Hence $\left(iii\right)\to a$.