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Question: A motorboat is racing towards the north at \(36km/h\) with respect to water and the speed of water c...

A motorboat is racing towards the north at 36km/h36km/h with respect to water and the speed of water current in that region is 10km/h10km/h towards east. The resultant speed of the boat is nearly

& A.37m/s \\\ & B.37km/h \\\ & C.46km/h \\\ & D.26km/h \\\ \end{aligned}$$
Explanation

Solution

Since the boat is moving on a river, which is flowing, then we can say that the velocity of the river acts as a hindrance for the movement of the boat. For the boat to cross the river, clearly it must overcome the velocity of the river. Then, we can calculate the resultant vector, from the parallelogram law of vector addition.
Formula used: vr=vb2+vw2v_{r}=\sqrt{v_{b}^{2}+v_{w}^{2}}

Complete step-by-step solution:
Let us consider that boat starts from a point OO as shown in the figure. Let the velocity of the boat be vb=36km/hv_{b}=36km/h and the velocity of the vw=10km/hv_{w}=10km/h.
Let us say that the boat from OO reaches the perpendicular point AA in the absence of the vwv_{w}. Due to the velocity of the river, vwv_{w}, clearly, this hinders the movement of the boat and thus the boat from OO reaches BB, instead of AA, which is inclined at an angle θ\theta with respect to O  AO\; A

Let vrv_{r} be the resultant vector. Then from the parallelogram law of vector addition, we can say that figure we can say that the resultant velocity vrv_{r}. Then the magnitude of the resultant vector is given as vr=vb2+vw2v_{r}=\sqrt{v_{b}^{2}+v_{w}^{2}}
    vr=362+102=1296+100=1390=37.36km/h\implies v_{r}=\sqrt{{36}^{2}+{10}^{2}}=\sqrt{1296+100}=\sqrt{1390}=37.36km/h
Then, we can say that the value of vr37km/hv_{r}\approx 37km/h
Then the direction of the resultant force is given by θ=tan1vwvr\theta=tan^{-1}\dfrac{v_{w}}{v_{r}}
Substituting, we get, θ=tan13610=tan13.6\theta=tan^{-1}\dfrac{36}{10}=tan^{-1}3.6
Thus the answer is B.37km/hB.37km/h

Note: This sum may seem very hard at first glance, but if one draws the diagram and labels the diagram correctly, it is easy to solve the question. The question uses only the basic parallelogram law of vector addition. Also the direction of the resultant force is given by θ=tan1vwvr\theta=tan^{-1}\dfrac{v_{w}}{v_{r}}, the direction is not asked here.