Question

A swimmer crosses the river along the line making an angle of $ 45{}^\circ $ with the direction of flow velocity of the river is $ 5{ m / sec} $ . Swimmer takes 6 seconds to cross the river of width $ 60{ m} $ . The velocity of the swimmer with respect to water will be:
(A) $ 10{ m/sec} $
(B) $ 12{ m/sec} $
(C) $ {5}\sqrt{5}{ m/sec} $
(D) $ 10\sqrt{5}{ m/sec} $

Answer 1

Hint: The velocity of the river can be assumed to be along x, y or z- axis (here we have taken y axis). As the swimmer crosses river making an angle, therefore the displacement can be found using the formula
$ {r=x\hat{i}}+{y\hat{j}} $
From here, the speed of the swimmer is found. The relative velocity is then found by subtracting the velocity of the river from the velocity of the swimmer. It’s magnitude gives us the required answer.

Complete step by step solution:
We assume that the river is flowing along y- axis. So, the velocity of river is given by
$ \overrightarrow{{{{v}}_{1}}}=5{\hat{j} m/sec} $
Now, the swimmer is crossing the river making an angle of $ 45{}^\circ $ with the direction of flow of the river. Therefore, using the equation
$ \overrightarrow{{r}}={x\hat{i}}+{y\hat{j}} $ , we get
Displacement $ =60{\hat{i}+60\hat{j}} $
 $ 60{ m} $ is the width of the river.
So, the speed of swimmer is
$ \overrightarrow{{{{v}}_{2}}}=\dfrac{{Displacement}}{{Time}} $
$ =\dfrac{60{\hat{i}+60\hat{j}}}{6} $
$ =10{\hat{i}+10\hat{j} m/sec} $
Hence, the velocity of swimmer relative to the river is given by
$ \overrightarrow{{{{v}}_{2}}}-\overrightarrow{{{{v}}_{1}}}=10{\hat{i}+}10{\hat{j}}-5{\hat{j}} $
$ =10{\hat{i}+}5{\hat{j} m/sec} $
Magnitude of velocity $ =\sqrt{{{\left( 10 \right)}^{2}}+{{\left( 5 \right)}^{2}}} $
$=\sqrt{100+25} $
$ =\sqrt{125} $
$ =5\sqrt{5}{ m/sec} $
So the correct option is (C).

Note:
The relative velocity of an object A with respect to another object B is equal to the resultant of the velocity of A and negative of the velocity of B.
Mathematically,
 $ {vAB=vA+}\left( -{vB} \right) $