1

Question

Consider a river 500 m wide. Speed of river water is 5 km/hr. Swimmers speed in still water is 3 km/hr. If swimmer wants to reach a point directly opposite the point from where he started to swim, find minimum distance he has to walk.

Open in App

Solution

Velocity of man, vm=3 km/h

BD = horizontal velocity of resultant R

Rx=5+3cosθ

t=(0.5)3sinθ

Now Hx=BD=(5+3cosθ)(0.53sinθ)

=5+3cosθ6sinθ

H to be minimum

dHdθ=0

ddθ(5+3cosθ6sinθ)=0

6sinθ(3(−sinθ))−(5+3cosθ)(6cosθ)=0

−18sin2θ−30cosθ−18cos2θ2θ=0

−18(sin2θ+cos2θ)−30cosθ=0

−18−30cosθ=0

cosθ=−1830

=−35

sinθ=√1−cos2θ

=45

H=5+3cosθ6sinθ

=5+3(−35)6×45

=1624=23 km

BD = horizontal velocity of resultant R

Rx=5+3cosθ

t=(0.5)3sinθ

Now Hx=BD=(5+3cosθ)(0.53sinθ)

=5+3cosθ6sinθ

H to be minimum

dHdθ=0

ddθ(5+3cosθ6sinθ)=0

6sinθ(3(−sinθ))−(5+3cosθ)(6cosθ)=0

−18sin2θ−30cosθ−18cos2θ2θ=0

−18(sin2θ+cos2θ)−30cosθ=0

−18−30cosθ=0

cosθ=−1830

=−35

sinθ=√1−cos2θ

=45

H=5+3cosθ6sinθ

=5+3(−35)6×45

=1624=23 km

0

View More

Join BYJU'S Learning Program

Join BYJU'S Learning Program