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Question: A steamer is moving due east with 10Km/h. To a man in the steamer, the wind appears to blow at \(5Km...

A steamer is moving due east with 10Km/h. To a man in the steamer, the wind appears to blow at 5Km/h5Km/h due north. Find the velocity of the wind.
(A) 55ms15\sqrt 5 m{s^{ - 1}} , 26.56{26.56^ \circ } North of east
(B) 5ms15m{s^{ - 1}} , 53.13{53.13^ \circ } North of east
(C) 55ms15\sqrt 5 m{s^{ - 1}} , 53.13{53.13^ \circ } North of east
(D) 5ms15m{s^{ - 1}} , 26.56{26.56^ \circ } North of east

Explanation

Solution

- A vector can be resolved into its components. This problem can be solved by resolving components of the velocities of the two ships. Equating components wherever possible may give us the required answer.

Step-by-step solution
If we take east as positive x-direction, North as positive y-direction. Then
The velocity of the steamer with respect to the Earth
vS/E=10i^{v_{S/E}} = 10\hat i
And the velocity of wind with respect to the steamer
vW/S=5j^{v_{W/S}} = 5\hat j
Thus, the velocity of wind with respect to the Earth
vW/E=vW/S+vS/E{v_{W/E}} = {v_{W/S}} + {v_{S/E}}
vW/E=10i^+5j^\Rightarrow {v_{W/E}} = 10\hat i + 5\hat j
The magnitude of the velocity of the wind
=102+52= \sqrt {{{10}^2} + {5^2}}
=55= 5\sqrt 5
And the angle between them is
tanθ=510=12θ=26.56\tan \theta = \dfrac{5}{{10}} = \dfrac{1}{2} \Rightarrow \theta = {26.56^ \circ }
I.e. wind is blowing at 55ms15\sqrt 5 m{s^{ - 1}} , 26.56{26.56^ \circ } north of east.
The correct answer is (A) 55ms15\sqrt 5 m{s^{ - 1}} , 26.56{26.56^ \circ } North of east

Additional Information
The magnitude of a vector can be calculated from its components using the Pythagoras theorem. If a vector has components vx{v_x} and vy{v_y} then the magnitude of vector V is given by the following expression:
V=vx2+vy2V = \sqrt {{v_x}^2 + {v_y}^2}
We can also resolve components of a vector in 3 dimensions. In the case of 3-dimensional representation, the components of V would be vx,{v_{x,}} , vy{v_y} and vz{v_z} . The magnitude of V in this case is given by modified Pythagoras Theorem as the following expression:
V=vx2+vy2+vz2V = \sqrt {{v_x}^2 + {v_y}^2 + {v_z}^2}

Note The components of a vector are constructed using trigonometric identities which satisfy the Pythagoras theorem for the magnitude of the vector. The components signify the directions in which a part of the main vector can be distributed and we designate those directions to be along the coordinate axes.