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A river is flowing due east with a speed of 3m/s. A swimmer can swim in still water at a speed of 4 m/s (Fig. 4.8).

(a) If the swimmer starts swimming due north, what will be his resultant velocity (magnitude and direction)?

(b) If he wants to start from point A on the south bank and reach opposite point B on the north bank,

(i) which direction should he swim?

(ii) what will be his resultant speed?

(c) From two different cases as mentioned in (a) and (b) above, in which case will he reach the opposite bank in a shorter time?

Answers (1)

a) If the swimmer starts swimming due north, what will be his resultant velocity

$V_{s}=4\; m/s$ due north

$V_{r}=4\; m/s$ due east

Now since both directions are perpendicular,

$\left | V_{r} \right |^{2}=4^{2}+3^{2}=5\; m/s$

$\tan \; \theta =\frac{V_{r}}{V_{s}}=0.75=36^{o}54'$ in the North direction

b) The swimmer wants to start from point A on the south bank and reach the opposite point B on the north bank

The swimmer makes an angle $\theta$ with the north.

From the figure we have the relation,

$V^{2}=v{_{s}}^{2}-v{_{r}}^{2}=16-9=7$

Henece $v=\sqrt{7}\; m/s$

Now we calculate the value of θ through the below formula,

$\tan \theta =\frac{v_{r}}{v}=\frac{3\sqrt{7}}{7}=1.13$

So, $\theta =48^{o}29'30''$ in the direction from North to West

c) we need to find from the above two scenarios that for the swimmer to reach the opposite bank in a shorter time

we know that the velocity component perpendicular to the river is 4m/s

let us assume the width of the river to be ‘w’

Time taken - North

$\frac{w}{4}=t1$

Time is taken in part b) when $v=\sqrt{7}\; m/s$

$\frac{w}{\sqrt{7}}=t2$

taking ratio,

$\frac{t1}{t2}=\frac{(\frac{w}{4})}{(\frac{w}{\sqrt{7}})}$

$4\; t1=\sqrt{7}\; t2$

Now as, $4>\sqrt{7}$

$t1<t2$

So, the swimmer will take a shorter time in case (a)

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