Question

a man crosses the river perpendicular to the river flow in time T seconds and travel and equal distance down the stream in capital T seconds the ratio of maths speed in still water to the speed of river water will be

Answer 1

Here's the solution.

Answer 2

Answer:

The answer is $\bold{\frac{v}{u}=\frac{t^{2}+T^{2}}{t^{2}-T^{2}}}$

Solution:

The time taken to ‘cross the river’ perpendicular to the river flow is  

$t=\frac{d}{\sqrt{v^{2}-u^{2}}}$

The time taken to cross equal distance down the stream is $T=\frac{d}{v+u}$

So the ratio of the both the time will give us  

$\frac{t}{T}=\frac{\left(\frac{d}{\sqrt{v^{2}-u^{2}}}\right)}{\left(\frac{d}{v+u}\right)}$

$\frac{t}{T}=\frac{v+u}{\sqrt{v^{2}-u^{2}}}$

Squaring on both sides, we get

$\left(\frac{t}{T}\right)^{2}=\left(\frac{v+u}{\sqrt{v^{2}-u^{2}}}\right)^{2}$

$\left(\frac{t}{T}\right)^{2}=\frac{v^{2}+u^{2}+2 u v}{v^{2}-u^{2}}$

Adding 1 on both sides, we get

$1+\left(\frac{t}{T}\right)^{2}=\frac{v^{2}+u^{2}+2 u v}{v^{2}-u^{2}}+1$

$\frac{T^{2}+t^{2}}{T^{2}}=\frac{v^{2}+u^{2}+2 u v+v^{2}-u^{2}}{v^{2}-u^{2}}$

$\frac{T^{2}+t^{2}}{T^{2}}=\frac{v^{2}+2 u v+v^{2}}{v^{2}-u^{2}}=\frac{2 v^{2}+2 u v}{(v-u)(v+u)}$

$\frac{T^{2}+t^{2}}{T^{2}}=\frac{2 v(v+u)}{(v-u)(v+u)}=\frac{2 v}{v-u}$

Taking reciprocal of it,

$\frac{2 T^{2}}{T^{2}+t^{2}}=\frac{v-u}{v}$

Subtracting 1 on either side, we get

$\Rightarrow 1-\frac{v-u}{v}=1-\frac{2 T^{2}}{T^{2}+t^{2}}$

$\Rightarrow \frac{v-v+u}{v}=\frac{T^{2}+t^{2}-2 T^{2}}{T^{2}+t^{2}}$

$\Rightarrow \frac{u}{v}=\frac{t^{2}-T^{2}}{T^{2}+t^{2}}$

$\bold{\Rightarrow \frac{v}{u}=\frac{t^{2}+T^{2}}{t^{2}-T^{2}}}$