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

Eshan
2 years ago
Praveen
13 Points
2 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
one year ago

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

11 months ago
Dear student,

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