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Question: A motorboat is racing towards the north at 25 \(km{h^{ - 1}}\)and the water current in that region i...

A motorboat is racing towards the north at 25 kmh1km{h^{ - 1}}and the water current in that region is 10kmh1km{h^{ - 1}}in the direction of 600{60^0}east of south. The resultant velocity of the boat is:
A. 11kmh1km{h^{ - 1}}
B. 22kmh1km{h^{ - 1}}
C. 33kmh1km{h^{ - 1}}
D. 44kmh1km{h^{ - 1}}

Explanation

Solution

In order to solve this problem, we should know the concept of resultant velocity and draw the resultant velocity diagram to get an idea for the given question. Both velocities are perpendicular to each other.

Step by step answer:
Given,
Velocity of the water current vc{v_c}​=10 km/s

Velocity of the motorboat vb{v_b}​=25km/h

Let VbandVc\overrightarrow {{V_b}} and\overrightarrow {{V_c}} be the velocities of boat and water respectively. V\overrightarrow V Be the resultant velocity of the boat.

Vb\overrightarrow {{V_b}} Is the velocity of boat heads towards the north and water velocity Vc\overrightarrow {{V_c}} is in the direction of 600{60^0}east of south.

Therefore, the angle between both the velocitiesVbandVc\overrightarrow {{V_b}} and\overrightarrow {{V_c}} is1200 {120^0}.

Angle between north and south east is 1200{120^0}

We know that the Resultant velocity is given by

VR=Vb2+Vc2+2VbVccos1200{V_R} = \sqrt {{V_b}^2 + {V_c}^2 + 2{V_b}{V_c}\cos {{120}^0}}

=252+102+2×25×10×(12)\sqrt {{{25}^2} + {{10}^2} + 2 \times 25 \times 10 \times \left( {\frac{{ - 1}}{2}} \right)}
=22kmh1km{h^{ - 1}}

Thus the resultant velocity of the boat is 22kmh1km{h^{ - 1}}
Hence the correct option is B

Note: The resultant is the vector sum of two or more vectors. It is the result of adding two or more vectors together. If two or more velocity vectors are added, then the result is a resultant velocity