Question

A boat can go across a lake and return in time ${{T}_{o}}$at a speed$\upsilon $. On a rough day there is a uniform current at speed ${{\upsilon }_{1}}$to help the onward journey and impede the return journey. If the time taken to go across and return on the same day be$T$, then $\dfrac{T}{{{T}_{o}}}$will be

A.)$\dfrac{1}{\left( 1-\dfrac{{{\upsilon }_{1}}^{2}}{{{\upsilon }^{2}}} \right)}$
B.)$\dfrac{1}{\left( 1+\dfrac{{{\upsilon }_{1}}^{2}}{{{\upsilon }^{2}}} \right)}$
C.)$\dfrac{1}{\left( 1-\dfrac{{{\upsilon }_{1}}^{2}}{{{\upsilon }^{2}}} \right)}$
D.)$\left( 1+\dfrac{{{\upsilon }_{1}}^{2}}{{{\upsilon }^{2}}} \right)$

Answer 1

Hint: If the speed of boat in still water is $\upsilon $ and speed of current is${{\upsilon }_{1}}$then, speed upstream$=\upsilon -{{\upsilon }_{1}}$ and speed downstream$=\upsilon +{{\upsilon }_{1}}$.

Formula used:

\[\text{Time = }\dfrac{\text{Distance}}{\text{speed}}\]
Speed upstream$=\upsilon -{{\upsilon }_{1}}$
Speed downstream$=\upsilon +{{\upsilon }_{1}}$

Complete step by step answer:
When a boat travels in the same direction as that of stream, we say it is travelling downstream and when boat travels in the direction opposite to that of stream, we say it is travelling upstream. In simple words, the direction along the water current is downstream and direction against water current is upstream.

To find the effective value of speed of boat in downstream we add the speed of current in speed of boat while in upstream we subtract the speed of current from speed of boat.
Let’s take the length of the lake as $L$, double of which, is also the distance travelled by the boat in both the cases.

For normal days, when water current is zero, we have ${{T}_{o}}=\dfrac{2L}{\upsilon }$, which is the time taken by boat to go across a lake and come back.

For rough day we have $T=\dfrac{L}{\upsilon +{{\upsilon }_{1}}}+\dfrac{L}{\upsilon -{{\upsilon }_{1}}}$

Where $\dfrac{L}{\upsilon +{{\upsilon }_{1}}}$ is the time taken by boat to travel downstream and $\dfrac{L}{\upsilon -{{\upsilon }_{1}}}$ is the time taken by boat to travel upstream.
$T=\dfrac{2L\upsilon }{{{\upsilon }^{2}}-{{\upsilon }_{1}}^{2}}$
Dividing $T$ and ${{T}_{o}}$ we get,

\[\begin{align}
  & \dfrac{T}{{{T}_{o}}}=\dfrac{2L\upsilon }{{{\upsilon }^{2}}-{{\upsilon }_{1}}^{2}}\times \dfrac{\upsilon }{2L}=\dfrac{{{\upsilon }^{2}}}{{{\upsilon }^{2}}-{{\upsilon }_{1}}^{2}} \\
 & \dfrac{T}{{{T}_{o}}}=\dfrac{1}{1-\left( \dfrac{{{\upsilon }_{1}}^{2}}{{{\upsilon }^{2}}} \right)} \\
\end{align}\]

Hence, the correct option is A.

Note:
Students should keep in mind that water current always opposes the movement of boats going upstream while it always supports the movement of boats going downstream. Always remember to subtract the speed of water current from the speed of the boat while moving upstream.