A steamer is moving due east with 10Km/h. To a man in the steamer, the wind appears to blow at $5Km/h$ due north. Find the velocity of the wind.
(A) $5\sqrt 5 m{s^{ - 1}}$ , ${26.56^ \circ }$ North of east
(B) $5m{s^{ - 1}}$ , ${53.13^ \circ }$ North of east
(C) $5\sqrt 5 m{s^{ - 1}}$ , ${53.13^ \circ }$ North of east
(D) $5m{s^{ - 1}}$ , ${26.56^ \circ }$ North of east
Answer
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Hint: - 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
${v_{S/E}} = 10\hat i$
And the velocity of wind with respect to the steamer
${v_{W/S}} = 5\hat j$
Thus, the velocity of wind with respect to the Earth
${v_{W/E}} = {v_{W/S}} + {v_{S/E}}$
$ \Rightarrow {v_{W/E}} = 10\hat i + 5\hat j$
The magnitude of the velocity of the wind
$ = \sqrt {{{10}^2} + {5^2}} $
$ = 5\sqrt 5 $
And the angle between them is
$\tan \theta = \dfrac{5}{{10}} = \dfrac{1}{2} \Rightarrow \theta = {26.56^ \circ }$
I.e. wind is blowing at $5\sqrt 5 m{s^{ - 1}}$ , ${26.56^ \circ }$ north of east.
The correct answer is (A) $5\sqrt 5 m{s^{ - 1}}$ , ${26.56^ \circ }$ 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 ${v_x}$ and ${v_y}$ then the magnitude of vector V is given by the following expression:
$V = \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 ${v_{x,}}$ , ${v_y}$ and ${v_z}$ . The magnitude of V in this case is given by modified Pythagoras Theorem as the following expression:
$V = \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.
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
${v_{S/E}} = 10\hat i$
And the velocity of wind with respect to the steamer
${v_{W/S}} = 5\hat j$
Thus, the velocity of wind with respect to the Earth
${v_{W/E}} = {v_{W/S}} + {v_{S/E}}$
$ \Rightarrow {v_{W/E}} = 10\hat i + 5\hat j$
The magnitude of the velocity of the wind
$ = \sqrt {{{10}^2} + {5^2}} $
$ = 5\sqrt 5 $
And the angle between them is
$\tan \theta = \dfrac{5}{{10}} = \dfrac{1}{2} \Rightarrow \theta = {26.56^ \circ }$
I.e. wind is blowing at $5\sqrt 5 m{s^{ - 1}}$ , ${26.56^ \circ }$ north of east.
The correct answer is (A) $5\sqrt 5 m{s^{ - 1}}$ , ${26.56^ \circ }$ 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 ${v_x}$ and ${v_y}$ then the magnitude of vector V is given by the following expression:
$V = \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 ${v_{x,}}$ , ${v_y}$ and ${v_z}$ . The magnitude of V in this case is given by modified Pythagoras Theorem as the following expression:
$V = \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.
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