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

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(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{

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

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(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:

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