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It can be easily shown that the time taken for covering the river is
t=d/vcos@
clearly for t(min), cos @ is max,ie, 1....
hence angle is 0° at wich he sud swim for min time
a river is flowing from w to e at a speed m units.a man on the south bank of river ,capable of swimming at n units in still water,wants to swim across the river in shortest time.derive the expression to find out the direction of man
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UP 0 DOWN 0 0 7
7 Answers
1i thnk since shortest time, so direction is to be south to north i.e, θ = 0°
the man has to move in the direction shown by the arrow in the diagram i.e from s to n
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1how to show that t=d/cos@ ..... ?plz can u explain the derivation?
1bhai prateek mujhe yeh samjh nahin aa raha boat waali problem mein tum log itna confuse kahe ho rahe ho aaj kal???
TIME will be minimum WHEN COMPONENT OF VELOCITY ALONG THE WIDTH IS MAXIMUM....tht will be amximum when the SWIMMER SWIMS PERPENDICULAR TO BANK!!!
PRATEEK FOR PROOF CONCENTRATE ON WORD """"COMPONENT"""""
1let's see the proof :
\vec{v_{r}} = absolute velocity of river
\vec{v_{br}} = velo. of boat wrt river or velo. of boat in still water
and
\vec{v_{b}} = absolute velocity of boatman
\vec{v_{b}} = \vec{v_{r}} + \vec{v_{br}}
now we consider the components
\vec{v_{bx}} = \vec{v_{rx}} + \vec{v_{brx}}
= \vec{v_{r}} + {v_{br}}sin θ
\vec{v_{by}} = \vec{v_{ry}} + \vec{v_{bry}}
= 0 + {v_{br}}cos θ
time taken by boatman to cross river is t = d\vec{v_{by}} = d{v_{br}}cos θ
hence proved
1but tht much is not even needed to be done in this one if we just realize this by common sense for shortest time!