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Home » A river is flowing due east with a speed 3 m/s. A swimmer can swim in still water at a speed of 4 m/s. a) if swimmer starts swimming due north, what will be his resultant velocity? b) if he wants to start from point A on south bank and reach opposite point B on north bank, i) which direction should he swim? ii) what will be his resultant speed?
A river is flowing due east with a speed 3 m/s. A swimmer can swim in still water at a speed of 4 m/s. a) if swimmer starts swimming due north, what will be his resultant velocity? b) if he wants to start from point A on south bank and reach opposite point B on north bank, i) which direction should he swim? ii) what will be his resultant speed?
CBSE | Class 11 | Motion in a Plane | NCERT | Physics
A river is flowing due east with a speed 3 m/s. A swimmer can swim in still water at a speed of 4 m/s. a) if swimmer starts swimming due north, what will be his resultant velocity? b) if he wants to start from point A on south bank and reach opposite point B on north bank, i) which direction should he swim? ii) what will be his resultant speed?

Answer:

Given,

The river’s velocity, vr, is 3 meters per second.

vs = 4 m/s is the speed of the swimmer.

a) When the swimmer swims due north, the y-component will have a velocity of 4 meters per second and the x-component will have a velocity of 3 meters per second. As a function of tan, the following is the resultant velocity.

v=\sqrt{v_{r}^{2}+v_{s}^{2}}=\sqrt{(3)^{2}+(4)^{2}}=5 m / s

Here, θ = 37oN

b)

If the swimmer intends to reach the place that is exactly opposite him and that is with regard to tan, the following is the consequent speed of the swimmer.

v=vs2–vr2=(4)2–(3)2=7m/sv=\sqrt{v_{s}^{2}-v_{r}^{2}}=\sqrt{(4)^{2}-(3)^{2}}=\sqrt{7} m / s

So, \theta=\tan ^{-1}(3 / \sqrt{7}) of north

 

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