Kinematics Question 660
Question: Two boats A and B, move away from a buoy anchored at the middle of a river along the mutually perpendicular straight lines: the boat A along the river and the boat B across the river. Having moved off an equal distance from the buoy the boat returned. What is the ratio of times of motion of boats $ \frac{{\tau _{A}}}{{\tau _{B}}}, $ if the velocity of each boat with respect to water is 1.2 times greater than the stream velocity?
Options:
A)2.3
B)1.8
C)0.5
D) 0.2
Show Answer
Answer:
Correct Answer: B
Solution:
[b] Suppose the stream velocity is $ {{\text{v}} _{\text{s}}}\text{=v} $ , then the velocity of each boat with respect to water is $ {{v _{b}}}=1.2\text{ v} $ . Let each boat travel a distance $ \ell $ .
Then for boat A, time of motion $ {\tau _{A}}=\frac{\ell }{v _{b}+v _{s}}+\frac{\ell }{v _{b}-v _{s}} $
$ =[ \frac{\ell }{1.2v+v}+\frac{\ell }{1.2v-v} ]=\frac{60\ell }{11v} $ ?..(i).
For the boat B, time of motion $ {\tau _{B}}=\frac{\ell }{\sqrt{v _{b}^{2}-v _{s}^{2}}}+\frac{\ell }{\sqrt{v _{b}^{2}-v _{s}^{2}}}=\frac{2\ell }{\sqrt{v _{b}^{2}-v _{s}^{2}}} $ -(ii)
$ =\frac{2\ell }{\sqrt{{{( 1.2v )}^{2}}-v^{2}}}=\frac{3.01\ell }{v} $ The ratio $ \frac{{\tau _{A}}}{{\tau _{B}}}=\frac{( 60\ell /11v )}{( 3.01\ell /v )}\approx 1.8 $
