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.