Question

To cyclist travelling due north along a straight road at 10km/h,the wind appears to come from the east.When he increases his speed to 15km/h,the wind appears to come from the direction N67°E.Find the speed and the direction of the wind over the ground.

Answer 1

The initial velocity of cyclist is

$$\vec V_{C1}=10\vec j$$

The wind appears to come from the east.

Suppose the relative velocity of the wind is

$$\vec V_{WC_1}=-a\vec i+0\vec j, a>0$$

The velocity of the wind is

$$\vec V_W=\vec V_{C1}+\vec V_{WC_1}=-a\vec i+10\vec j$$

 Cyclist increases his speed to 15km/h

$$\vec V_{C2}=15\vec j$$

The wind appears to come from the direction N67°E.

Suppose the relative velocity of the wind is

$$\vec V_{WC2}=-c\sin(67\degree)\vec i-c\cos(67\degree)\vec j, c>0, d>0$$

The velocity of the wind is

$$\vec V_W=\vec V_{C2}+\vec V_{WC_2}$$

$$=-c\sin(67\degree)\vec i+(15-c\cos(67\degree))\vec j$$

Then

$$\begin{matrix} a =c\sin(67\degree) \\ 10=15-c\cos(67\degree) \end{matrix}$$

$$c=\dfrac{5}{\cos(67\degree)}$$

$$a=5\tan(67\degree)$$

$$\vec V_W=-5\tan(67\degree)\vec i+10\vec j$$

$$|\vec V_W|=\sqrt{(-5\tan(67\degree))^2+(10)^2}$$

$$=5\sqrt{4+\tan^2(67\degree)}$$

$$\tan(\theta)=\dfrac{10}{-5\tan(67\degree)}$$

$$\theta=180\degree-\tan^{-1}(\dfrac{2}{\tan(67\degree)})$$

$$|\vec V_W|\approx15.45\ km/h$$

$$\theta\approx139.67\degree$$