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To cross the river in shortest distance, a swimmer should swim making angle theta with the upstream. What is the ratio of the time taken to swim across in the shortest time to that in swimming across over shortest distance. [Assume speed of swimmer in still water is greater than the speed of river flow]

To cross the river in shortest distance, a swimmer should swim making angle theta  with the upstream. What is
the ratio of the time taken to swim across in the shortest time to that in swimming across over shortest
distance. [Assume speed of swimmer in still water is greater than the speed of river flow]

 

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Grade:12th pass

4 Answers

Eshan
askIITians Faculty 2095 Points
5 years ago
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Praveen
13 Points
4 years ago
t(shortest)=t1=d/v(sr)
t(shortest distance)=t2=d/v(sr)sin(theta)
=t1/t2= sin (theta)
Where d is width of river
Tejaswi
15 Points
4 years ago
Answer is sin (theta)
 
As t shortest=t¹=d/v(sr)
T(shortest)=t²=d/v(sr) sin (theta)
=t¹/t²=sin (theta)
Where d is width of river
 
Kushagra Madhukar
askIITians Faculty 628 Points
3 years ago
Dear student,
Please find the solution to your problem.
 
Let the velocity of river be vr and velocity of swimmer be vsr
 
For the shortest time case-
Time taken to cross river, t1 = d/vsr
 
For shortest distance case-
Time taken to cross river, t2 = d/vsrcosθ
where θ is the angle made by swimmer w.r.t the direction normal to the direction of flow of river.
where cosθ = (vsr2 – vr2)1/2 /vsr
 
Hence, t1/t2 = cosθ = √(1 – (vr/vsr)2)
 
Thanks and regards,
Kushagra
 

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