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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.

7 years ago


Answers : (1)


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


7 years ago

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