Hey there! We receieved your request
Stay Tuned as we are going to contact you within 1 Hour
One of our academic counsellors will contact you within 1 working day.
Click to Chat
1800-5470-145
+91 7353221155
Use Coupon: CART20 and get 20% off on all online Study Material
Complete Your Registration (Step 2 of 2 )
Sit and relax as our customer representative will contact you within 1 business day
OTP to be sent to Change
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.
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/va-Vr.total 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 =Vb.to 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
Register Yourself for a FREE Demo Class by Top IITians & Medical Experts Today !