A motorboat is racing towards north at 25km/h and the water current in that region is 10km/h in the direction of 60∘ east of south. The resultant velocity of the boat is
A
22km/hr, at an angle tan−1(0.343) North of East.
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B
22km/hr, at an angle tan−1(0.433) East of North.
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C
20km/hr, at an angle tan−1(0.433) East of North.
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D
20km/hr, at an angle tan−1(0.343) North of East.
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Solution
The correct option is B22km/hr, at an angle tan−1(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=21.8≈22km/hr Now, for direction we have tanϕ=vcsin(120∘)vb+vccos(120∘)=10sin(120∘)25+10cos(120∘) ⇒tanϕ=0.433⇒ϕ=tan−1(0.433) East of North.