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

Total Price: Rs.

There are no items in this cart.
Continue Shopping
Prasanjeet Mohanty Mohanty Grade: 12

two boats A and B move away from a buoy anchored at the middle of a river along the mutually perpendicular staright lines, A along the river & B across the river. having moved off an equal distance from the buoy, the boats returned.Find the ratio of times of motion of boats Ta/Tb if the velocity of each boat w.r.t water is n=1.2 times greater than the stream velocity.

8 years ago

Answers : (1)

piyush garg
9 Points

MOTION of boat A:at t=0, both boats are at o.boat a moves along stream velocity upto point A1.let Va is velocity of boat A and Vr is velocity of river

let t1 is time taken by boat A to travel a distance L in direction of river and t2 is time taken by boatA to return.then t1=L/va+vr and t2=L/ time Ta=L/Va+Vr+L/va-Vr

motion of boat B:boat B moves along perpendicular direction of river let  velocity make angle R with vertical let  velocity of boat B travel in perpendicular direction,it is necessary that component Vbsin(R) is balanced by Vr

Vr=Vbsin(R).resultant velocity of boat b in perpendicular direction is Vbcos(R).t1=L/Vbcos(R) and t2=L/Vbcos(R).total time=2L/Vbcos(R).Ta/Tb=L/va+vr+L/va-vr divided by 2L/Vbcos(R).put Vb=Va=nVr.

ta/tb=n/(n^2-1)^1/2=1.8 ans.

so this problem is based upon relative motion in two dimensions


8 years ago
Think You Can Provide A Better Answer ?
Answer & Earn Cool Goodies
  • Complete Physics Course - Class 12
  • OFFERED PRICE: Rs. 2,756
  • View Details
  • Complete Physics Course - Class 11
  • OFFERED PRICE: Rs. 2,968
  • View Details

Ask Experts

Have any Question? Ask Experts

Post Question

Answer ‘n’ Earn
Attractive Gift
To Win!!! Click Here for details