SATHEE: Kinematics Question 660

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 $\frac{τ_{A}}{τ_{B}},$ 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 $v_{s} =v$ , then the velocity of each boat with respect to water is $v_{b} = 1.2 v$ . Let each boat travel a distance $ℓ$ .

Then for boat A, time of motion $τ_{A} = \frac{ℓ}{v_{b} + v_{s}} + \frac{ℓ}{v_{b} - v_{s}}$

$= [\frac{ℓ}{1.2 v + v} + \frac{ℓ}{1.2 v - v}] = \frac{60 ℓ}{11 v}$ ?..(i).

For the boat B, time of motion $τ_{B} = \frac{ℓ}{\sqrt{v_{b}^{2} - v_{s}^{2}}} + \frac{ℓ}{\sqrt{v_{b}^{2} - v_{s}^{2}}} = \frac{2 ℓ}{\sqrt{v_{b}^{2} - v_{s}^{2}}}$ -(ii)

$= \frac{2 ℓ}{\sqrt{{(1.2 v)}^{2} - v^{2}}} = \frac{3.01 ℓ}{v}$ The ratio $\frac{τ_{A}}{τ_{B}} = \frac{(60 ℓ / 11 v)}{(3.01 ℓ / v)} \approx 1.8$

Kinematics Question 660

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Kinematics Question 660

Options:

Answer:

Solution:

Engineering Preparation

Medical Preparation

NCERT Books & Solutions

Important Resources

Other Resources

Syllabus

Test Platform

Sample Mock Test

Mentorship

NCERT Exercise Video Solution

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