466sorry ...
the first line wud b :
2 boats A and B , move away from a buoy anchored at the middle of a river along mutually perpendicular straight lines : the boat A along the river ... the boat B .....
To both, A and B, move away from a buoy and ancurd at the middle of a river, and the boat 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 Ï„A Ï„B . If the velocity of each boat with respect to water is 1. 2 times greater than the stream velocity
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2 Answers
283Let boats speed be V and stream speed be U and distance be S
For boat A total time takes is given by t1= S/(v+u) for donstream and t2 = S/v-u for upstream
Ta= t1 + t2 = 2SV/ v2 - u2
For Boat B its quite complicated but still...
B has to Perpendicular so it has to make an angle x such that
Vsinx= U ....sinx= U/V
we did this one in river boat problem in cases where drift has to be minimum
now the cos component of V will help it moving forward.
Vcosx=√V2 - U2 (by trigo)
Now since in perpedicualr direction the speed remains const through out so
Tb= 2S/√V2 - U2
Ta/Tb= V/√V2 - U2
= 1.2/√1.22 - 1
=1.8
- Himanshu Giria thanks...Upvote·0· Reply ·2014-02-09 03:07:21
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