Thank you for registering.

One of our academic counsellors will contact you within 1 working day.

Please check your email for login details.
MY CART (5)

Use Coupon: CART20 and get 20% off on all online Study Material

ITEM
DETAILS
MRP
DISCOUNT
FINAL PRICE
Total Price: Rs.

There are no items in this cart.
Continue Shopping

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]

 

Question Image
Grade:12th pass

4 Answers

Eshan
askIITians Faculty 2095 Points
2 years ago
568-398_123.png
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
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 629 Points
11 months 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
 

Think You Can Provide A Better Answer ?

Provide a better Answer & Earn Cool Goodies See our forum point policy

ASK QUESTION

Get your questions answered by the expert for free