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Question

A motorboat is racing towards north at 25 km/h and the water current in that region is 10 km/h in the direction of 60 east of south. The resultant velocity of the boat is

A
22 km/hr, at an angle tan1(0.343) North of East.
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B
22 km/hr, at an angle tan1(0.433) East of North.
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C
20 km/hr, at an angle tan1(0.433) East of North.
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D
20 km/hr, at an angle tan1(0.343) North of East.
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Solution

The correct option is B 22 km/hr, at an angle tan1(0.433) East of North.
The vector vb representing the velocity of the motorboat and the vector vc representing the water current are shown in the figure in directions specified by the problem.


Using the parallelogram method of addition, the resultant R is obtained in the direction shown in the figure.
So, the magnitude of the resultant is given by
R=v2b+v2c+2vbvccosθ, where θ=120
R=252+102+2×25×10cos(120)
R=21.822 km/hr
Now, for direction we have
tanϕ=vcsin(120)vb+vccos(120)=10sin(120)25+10cos(120)
tanϕ=0.433 ϕ=tan1(0.433) East of North.

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