Question

Wind is blowing from the south at $5m{s}^{-1}$. To a cyclist it appears to blowing from the east at $5m{s}^{-1}$. Find the velocity of the cyclist.

Answer 1

Since the velocity of the cyclist is a vector, it can be resolved into components.

Since s/he doesn't feel any component of wind from the north or north-east direction, it means that his/her northern component of velocity is equal to the wind's velocity, =5m/s

Since s/he feels wind = 5m/s from east, his/her component towards east is 5m/s.

By parallelogram law of vector addition, we have $A=\sqrt{N^2+E^2}$

Here A is actual velocity, and N and E are the respective components.

Hence $A=5\sqrt{2}$ $m/s$

Let his/her angle with East be $\alpha$.

$\alpha =ta{n}^{-1}\left(\frac{N}{E}\right)=ta{n}^{-1}1={45}^{{}^O}$

The cyclist is moving with a velocity$=5\sqrt{2}$ m/s exactly towards north-east.