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
Get your questions answered by the expert for free
You will get reply from our expert in sometime.
We will notify you when Our expert answers your question. To View your Question
Win Gift vouchers upto Rs 500/-
Register Yourself for a FREE Demo Class by Top IITians & Medical Experts Today !