Question

A man travelling east at 8 km per hour finds that the wind seems to blow directly from the north. On doubling the speed, he finds that it appears to come from N-E. Find the velocity of the wind and its direction.
A. $8\sqrt 2$ , its direction is from N-S
B. $8\sqrt 2$ , its direction is from N-W
C. $8\sqrt 2$ , its direction is from S-W
D.None of these

Answer 1

Hint : We are going to solve this question in two cases. In the first case the speed of man is 8 and wind is blowing from north and in second case speed of man is 16 and wind blows from N-E. For both these cases, we need to suppose two unit vectors $\mathop i\limits^ \wedge$ and $\mathop j\limits^ \wedge$ for north and east. Solve both the cases separately and combine the results to find the velocity of wind and its direction.

Complete step by step solution:

Velocity of wind relative to man = Actual velocity of wind – Actual velocity of man - - - - - - (1)
Let $\mathop i\limits^ \wedge$ and $\mathop j\limits^ \wedge$ represent unit vectors along the east and north direction.
Actual velocity of wind $= x\mathop i\limits^ \wedge + y\mathop j\limits^ \wedge$
During the first case, the man is moving at 8 km per hour in the east direction.
Velocity of man $= 8\mathop i\limits^ \wedge$
And at that time, the wind seemed to blow from north.
Velocity of wind relative to man $= - p\mathop j\limits^ \wedge$
Putting all these values in equation (1), we get
$\- p\mathop j\limits^ \wedge = x\mathop i\limits^ \wedge + y\mathop j\limits^ \wedge - 8\mathop i\limits^ \wedge \\\ \- p\mathop j\limits^ \wedge = \left( {x - 8} \right)\mathop i\limits^ \wedge + y\mathop j\limits^ \wedge \;$
On comparing,
$\left( {x - 8} \right) = 0$ and $y = - p$
$\Rightarrow x = 8$ and $y = - p$
Now, during second case, then man is doubling his speed and he thinks the wind seems to blow from N-E direction.
Velocity of man $= 16\mathop i\limits^ \wedge$
Velocity of wind relative to man$= - k\left( {\mathop i\limits^ \wedge + \mathop j\limits^ \wedge } \right)$
Putting these values in equation (1), we get

$$\- k\left( {\mathop i\limits^ \wedge + \mathop j\limits^ \wedge } \right) = x\mathop i\limits^ \wedge + y\mathop j\limits^ \wedge - 16\mathop i\limits^ \wedge \\\ \- k\left( {\mathop i\limits^ \wedge + \mathop j\limits^ \wedge } \right) = \left( {x - 16} \right)\mathop i\limits^ \wedge + y\mathop j\limits^ \wedge \\\ \- k\mathop i\limits^ \wedge - k\mathop j\limits^ \wedge = \left( {x - 16} \right)\mathop i\limits^ \wedge + y\mathop j\limits^ \wedge \\\$$

On comparing, we get,
$\left( {x - 16} \right) = - k$ and $y = - k$
Now, we know that $x = 8$
$\- k = 8 - 16 \\\ \- k = - 8 \\\ k = 8 \;\$
$y = - 8$
Therefore, velocity of wind $= x\mathop i\limits^ \wedge + y\mathop j\limits^ \wedge$

$$= 8\mathop i\limits^ \wedge - 8\mathop j\limits^ \wedge \\\ = 8\left( {\mathop i\limits^ \wedge \mathop { - j}\limits^ \wedge } \right) \\\$$

Therefore, magnitude of vector $= \sqrt {{x^2} + {y^2}}$
$= \sqrt {{8^2} + {8^2}} \\\ = \sqrt {128} \\\ = 8\sqrt 2 \;$
Now to find the direction of the wind,
$\tan \theta = - \dfrac{x}{y} \\\ \tan \theta = - \dfrac{8}{8} \\\ \tan \theta = - 1 \\\ \theta = - 45^\circ \;$
Therefore, the wind is blowing at velocity of $8\sqrt 2$ and in N-W direction.
So, the correct answer is option B.
So, the correct answer is “Option B”.

Note : Notice that in first case the velocity of wind is $- p\mathop j\limits^ \wedge$ . The negative sign is because of the arrow coming to the origin as shown in the graph above. Similarly, in the second case the velocity of wind is$- k\left( {\mathop i\limits^ \wedge + \mathop j\limits^ \wedge } \right)$. Here too, the arrow is coming towards origin as shown in the graph.