Question

The wind appears to blow from north to a man moving in the north-east direction. When he doubles his velocity, the wind appears to move in the direction cot${}^{-1}$ (2) east of north. If the actual magnitude of velocity of the wind is $V=\frac{1}{\sqrt{x}}\times$ magnitude of velocity of man and direction along east. Find $x$.

Answer 1

Suppose the wind velocity is v

Here, OA represent the magnitude and direction of man

OC represent the relative speed of wind

If we produce AO to D such that OD=OA=u

OD represent the the velocity of the man in reverse direction

Now, we complete the parallelogram OBCD. OB represents the actual velocity of the wind

suppose the angle BOE= $\theta$ and angle OCD= 90 - $\theta$

In triangle COD,

$\frac{OD}{\sin (90-\theta )}$ = $\frac{CD}{\sin \left(45\right)}$

$\frac{u}{v}=\sqrt{2}\cos \theta$ $⟶$ (1)

by calculation we get cos$\theta$ = 1

$\frac{v}{u}=\sqrt{2}$

u=$\frac{v}{\sqrt{2}}$

so answer is x=2