A man has to reach point B separated by water.He can swim straight from A to B or get to shore at some point and run from there.He can swim at a speed of 10 m/s on water and can run at 15m/s on shore.What would be the total distance he has to cover so as to reach point B at the earliest?
Answer:
The resultant u ad v should be along the line AB.
V = velocity of Man
\begin{lgathered}V_x=u-Vsin\theta\\ V_y=Vcos\theta\\ tan (45deg) = V_y /V_x\\1=Vcos\theta/u-Vsin\theta\end{lgathered}
V
x
=u−Vsinθ
V
y
=Vcosθ
tan(45deg)=V
y
/V
x
1=Vcosθ/u−Vsinθ
Hence, V=u/sin\theta+cos\theta=u/\sqrt{2}sin(\theta+45)V=u/sinθ+cosθ=u/
2
sin(θ+45)
Since V is minimum at, θ + 45° = 90°
Therefore, V_{min} =u /\sqrt2V
min
=u/
2
Answer:
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