On a calm day,a boat can go across a lake and | vedclass

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

Question

(B) Let the distance across the lake be $S$.On a calm day,the speed of the boat is $V$. The time taken to go and return is $T_0 = \frac{S}{V} + \frac{

Vedclass · Accepted Answer

$\frac{1}{1 - v^2 / V^2}$

Vedclass · Answer

(B) Let the distance across the lake be $S$.On a calm day,the speed of the boat is $V$. The time taken to go and return is $T_0 = \frac{S}{V} + \frac{S}{V} = \frac{2S}{V}$.On a rough day,the current speed is $v$. The speed during the onward journey is $(V + v)$ and during the return journey is $(V - v)$.The total time taken is $T = \frac{S}{V + v} + \frac{S}{V - v}$.$T = S \left( \frac{V - v + V + v}{V^2 - v^2} \right) = \frac{2SV}{V^2 - v^2}$.Now,calculating the ratio $T / T_0$:$T / T_0 = \frac{2SV / (V^2 - v^2)}{2S / V} = \frac{V^2}{V^2 - v^2}$.Dividing the numerator and denominator by $V^2$,we get $T / T_0 = \frac{1}{1 - v^2 / V^2}$.

On a calm day,a boat can go across a lake and return in time $T_0$ at a speed $V$. On a rough day,there is a uniform current at speed $v$ that helps the onward journey and impedes the return journey. If the time taken to go across and return on the rough day is $T$,then $T / T_0$ is:

Similar Questions

$A$ train is moving towards the east and a car is moving towards the north,both with the same speed. The observed direction of the car to a passenger in the train is

$A$ boat is moving with a velocity of $3\hat i + 4\hat j$ in a river,and the water is moving with a velocity of $-3\hat i - 4\hat j$ with respect to the ground. The relative velocity of the boat with respect to the water is:

When a train is at rest,rain drops fall vertically. When it moves along east with a velocity equal to that of rain,the rain will appear to fall,for a passenger in the train

Vedclass Test Series

Exam Paper Generator

Online Exam Module