A swimmer crosses a river of width d flowing | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(C) Let $v_r$ be the velocity of the swimmer with respect to the river.The swimmer heads at an angle of $120^{\circ}$ with the river flow $v$. The com

Vedclass · Accepted Answer

$4(2-\sqrt{3})v$

Vedclass · Answer

(C) Let $v_r$ be the velocity of the swimmer with respect to the river.The swimmer heads at an angle of $120^{\circ}$ with the river flow $v$. The component of $v_r$ perpendicular to the river flow is $v_r \sin(120^{\circ}-90^{\circ}) = v_r \sin 30^{\circ} = v_r/2$ is incorrect based on the geometry. Actually,the angle with the normal to the river flow is $120^{\circ}-90^{\circ} = 30^{\circ}$.The time taken to cross the river is $t = \frac{d}{v_r \cos 30^{\circ}} = \frac{d}{v_r (\sqrt{3}/2)} = \frac{2d}{\sqrt{3} v_r}$.The drift in the direction of the river flow is given by $x = (v - v_r \sin 30^{\circ})t$.Given $x = d/2$,we have $d/2 = (v - v_r/2) \left( \frac{2d}{\sqrt{3} v_r} \right)$.Simplifying,$1/2 = \frac{2v - v_r}{\sqrt{3} v_r} \implies \sqrt{3} v_r = 4v - 2v_r$.$v_r(2 + \sqrt{3}) = 4v \implies v_r = \frac{4v}{2 + \sqrt{3}}$.Rationalizing the denominator: $v_r = \frac{4v(2 - \sqrt{3})}{(2 + \sqrt{3})(2 - \sqrt{3})} = \frac{4v(2 - \sqrt{3})}{4 - 3} = 4(2 - \sqrt{3})v$.

$A$ swimmer crosses a river of width $d$ flowing at velocity $v$. While swimming,he heads himself always at an angle of $120^{\circ}$ with the river flow and on reaching the other end he finds a drift of $d/2$ in the direction of flow of river. The speed of the swimmer with respect to the river is

Similar Questions

Two trains $A$ and $B$ travel on two parallel rail tracks in opposite directions with speeds $v_1$ and $v_2$, respectively. They take $4 \ s$ to pass each other at these speeds. If the speed of train $A$ is increased by $50 \%$, then they take $3 \ s$ to pass each other. The ratio $v_1 / v_2$ is:

Out of the two cars $A$ and $B$,car $A$ is moving towards the east with a velocity of $10 \, m/s$,whereas car $B$ is moving towards the north with a velocity of $20 \, m/s$. The velocity of $A$ with respect to $B$ is ......... $m/s$ (nearly).

$A$ ship $A$ is moving Westwards with a speed of $10 \, km \, h^{-1}$ and a ship $B$,$100 \, km$ South of $A$,is moving Northwards with a speed of $10 \, km \, h^{-1}$. The time after which the distance between them becomes shortest is ........ $hr$.

There are two particles $P$ and $Q$ separated by a distance of $10 \ km$ and moving with velocities of $10 \ kmph$ as shown in the figure. The time elapsed before they reach their minimum separation is.........$hr$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module