Question

Two boats P and Q move away from a buoy anchored at the middle of a river along the mutually perpendicular straight lines, the boat P along the river and the boat Q across the river. Having moved off an equal distance from the buoy the boats returned. If the velocity of each boat with respect to water is 1.2 times greater than the stream velocity find the ratio of times of motion of boats $\dfrac{{{T}_{P}}}{{{T}_{Q}}}$ after rounding off.

Answer 1

Hint: When a boat moves across a river, it experiences both upstream and downstream motion. Hence along with taking the velocity of the stream as well. The upstream velocity is to be added and the downstream velocity is to be subtracted.

Complete step by step solution:
We’re given that both the boats travel equal distances in perpendicular directions. Let the distance be d.
Consider the stream velocity be ${{V}_{R}}$
Relative velocity of Q with respect to river is the same as that of P’s relative velocity with respect to river. Let it be ${{V}_{r}}$
We’re also given that $\dfrac{{{V}_{r}}}{{{V}_{R}}}=1.2$

Forward motion of Q:
\[{{\overrightarrow{V}}_{Q/R}}={{\overrightarrow{V}}_{r}}=-{{V}_{r}}\sin \theta \hat{i}+{{V}_{r}}cos\theta \hat{j}\]
$\overrightarrow{{{V}_{Q}}}-\overrightarrow{{{V}_{R}}}=-{{V}_{r}}\sin \theta \overset{\scriptscriptstyle\frown}{i}+{{V}_{r}}\cos \theta \hat{j}$
$\overrightarrow{{{V}_{Q}}}=({{V}_{R}}-{{V}_{r}}\sin \theta )\overset{\scriptscriptstyle\frown}{i}+{{V}_{r}}\cos \theta \hat{j}$
Since Q moves vertically, its horizontal component is zero
That is,
 $
 {{V}_{R}}-{{V}_{r}}\sin \theta =0 \\
 \Rightarrow {{V}_{R}}={{V}_{r}}\sin \theta \\
 \therefore \sin \theta =\dfrac{{{V}_{R}}}{{{V}_{r}}}=\dfrac{1}{1.2} \\
$
If the time taken for Q to travel a distance d in forward motion is ${{T}_{Q}}$
Then total time,
 ${{T}_{Q}}=\dfrac{2d}{1.2{{V}_{R}}\cos \theta }$ (Same for backward motion as well)

Forward motion of P:
Velocity of P with respect to river
\[\overrightarrow{{{V}_{PR}}}={{V}_{r}}\hat{i}\] since the direction of velocity of both the river and the boat is same.
\[
  \overrightarrow{{{V}_{P}}}-\overrightarrow{{{V}_{R}}}={{V}_{r}}\hat{i} \\
 \Rightarrow \overrightarrow{{{V}_{P}}}=\overrightarrow{{{V}_{R}}}+{{V}_{r}}\hat{i} \\
 \Rightarrow \overrightarrow{{{V}_{P}}} =({{V}_{R}}+{{V}_{r}})\hat{i} \\
 \Rightarrow \overrightarrow{{{V}_{P}}} ={{V}_{R}}(1+1.2)\hat{i}=2.2{{V}_{R}}\hat{i} \\
\]
Time taken to complete forward motion by P:
${{T}^{'}}=\dfrac{d}{2.2{{V}_{R}}}$

Backward motion of P:
\[\overrightarrow{{{V}_{AR}}}=-{{V}_{r}}\hat{i}\]
\[
  \overrightarrow{{{V}_{P}}}-\overrightarrow{{{V}_{R}}}=-{{V}_{r}}\hat{i} \\
  \Rightarrow \overrightarrow{{{V}_{P}}}=\overrightarrow{{{V}_{R}}}-{{V}_{r}}\hat{i} \\
  =({{V}_{R}}-{{V}_{r}})\hat{i} \\
  ={{V}_{R}}(1-1.2)\hat{i}=-0.2{{V}_{R}}\hat{i} \\
\]
Time taken: ${{T}^{“}}=\dfrac{d}{0.2{{V}_{R}}}$
Therefore total time taken by P
$
  {{T}_{P}}=\dfrac{d}{2.2{{V}_{R}}}+\dfrac{d}{0.2{{V}_{R}}} \\
  {{T}_{P}} =\dfrac{d}{{{V}_{R}}}(\dfrac{1}{2.2}+\dfrac{1}{0.2})=\dfrac{5.45d}{{{V}_{R}}} \\
$
Hence, $\dfrac{{{T}_{P}}}{{{T}_{Q}}}=\dfrac{\dfrac{5.45d}{{{V}_{R}}}}{\dfrac{2d}{1.2{{V}_{R}}\cos \theta }}=\dfrac{5.45\times 1.2\cos \theta }{2}$
Since,
$
  \sin \theta =\dfrac{1}{1.2} \\
  \cos \theta =\dfrac{\sqrt{{{1.2}^{2}}-1}}{1.2}=0.36 \\
$
Hence,
$\dfrac{{{T}_{P}}}{{{T}_{Q}}}=\dfrac{5.45\times 1.2\times 0.36}{2}\approx 1.8$
Hence, the required ratio has been obtained.

Note: If the ratio between the relative velocity and the stream velocity is given and if it is supposedly $\eta $ then we can find the ratios of time of objects in perpendicular motion using the formula
\[\dfrac{{{T}_{1}}}{{{T}_{2}}}=\dfrac{\eta }{\sqrt{{{\eta }^{2}}-1}}\]