A swimmer wishes to cross a 12 km wide river flowing at 5 km/h. His speed with respect to water is 3 km/h. If he heads in a direction making an angle
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Solution :
Velocity of man vm=3km/h
BD= Horizontal distance for resultant velcoity R
X= component of resultant R=5+3cosθ
t=0.5/3sinθ
Which is same for horizontal component of velcoity
H=BD =(5+3cos θ)(0.5 /3) sinθ
=5+3 cosθ/6 sin θ
For H to be minimum (dH/dθ)=0
d/dθ(5+3cosθ/6sinθ)=0
18(sin²θ+cos²θ)-30cosθ=0
30Cosθ=18
cosθ=-18/30=-3/5
Sinθ=√1-Cos²θ=4/5
H=5+3cosθ/6sinθ
=5+3(-3/5)/6x4/5
=25-9/24
=16/24
=2/3 km
Velocity of man vm=3km/h
BD= Horizontal distance for resultant velcoity R
X= component of resultant R=5+3cosθ
t=0.5/3sinθ
Which is same for horizontal component of velcoity
H=BD =(5+3cos θ)(0.5 /3) sinθ
=5+3 cosθ/6 sin θ
For H to be minimum (dH/dθ)=0
d/dθ(5+3cosθ/6sinθ)=0
18(sin²θ+cos²θ)-30cosθ=0
30Cosθ=18
cosθ=-18/30=-3/5
Sinθ=√1-Cos²θ=4/5
H=5+3cosθ/6sinθ
=5+3(-3/5)/6x4/5
=25-9/24
=16/24
=2/3 km
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