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[SOLVED] proof of known velocity of wind

Dhamnekar Winod

Active member
Nov 17, 2018
163
1627226956610.png

My attempt to answer this question: Let the actual velocity of wind is $\vec{v}=x\hat{i} + y\hat{j}$ where $\hat{i}$ and $\hat{j} $ represents velocities of 1KM per hour towards east and north respectively. As the person is going northeast with a velocity of 6KM/hr, his actual velocity is $ 3\sqrt{2} \hat{i} +3\sqrt{2}\hat{j}$

Then the velocity of wind relative to person is $x\hat{i} -y\hat{j}- 3\sqrt{2}\hat{i} -3\sqrt{2}\hat{j}$ which is parallel to $-3\sqrt{2}\hat{j}$ as it appears to blow from the north. Hence $x=3\sqrt{2},y=(3\sqrt{2}-k)\hat{j}$

When the velocity of the person becomes $6\sqrt{2}\hat{i} + 6\sqrt{2}\hat{j} $, the velocity of the wind relative to person is $(3\sqrt{2}\hat{i}+(3\sqrt{2}-k)\hat{j}) - 6\sqrt{2}\hat{i} -6\sqrt{2}\hat{j} $ So velocity of wind relative to person is $-3\sqrt{2}\hat{i}+(-3\sqrt{2}-k)\hat{j}$

$\frac{-3\sqrt{2}-k}{-3\sqrt{2}}=2 \rightarrow k=3\sqrt{2}$, Hence velocity of wind is $3\sqrt{2}\hat{i} +(3\sqrt{2}-3\sqrt{2})\hat{j}$.

Thus we showed that actual velocity of wind is $3\sqrt{2}$ KM/hr towards east.
 
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