Solution

Engineering Mechanics (Dynamics)

Chapter 12

Exercise 222

The measurements at the second time indicate that ${v}'_c = 80$ km/h and the wind is coming from the north-east, then we have,

$$\bold{v}_w = 80\bold{j} - {v}'_w\cos45^\circ\bold{i} - {v}'_w\sin45^\circ\bold{j} = \bigg(80 - \frac{{v}'_w}{\sqrt{2}}\bigg)\bold{j} - \frac{{v}'_w}{\sqrt{2}}\bold{i}~~~~~(*)$$

The absolute velocity of the wind $v_w$ is the same for both measurements! This requires that,

$$50\bold{j} - v_w\bold{i} = (80 - \frac{{v}'_w}{\sqrt{2}})\bold{j} - \frac{{v}'_w}{\sqrt{2}}\bold{i}$$

Equating the $y$-components yields

$$50 = 80 - \frac{{v}'_w}{\sqrt{2}}$$

$${v}'_w = 30\sqrt{2}\mathrm{~km/h}$$

Substituting into Eq. (*) gives the velocity of the wind,

$$\bold{v}_w = \bigg(80 - \frac{30\sqrt{2}}{\sqrt{2}}\bigg)\bold{j} - \frac{30\sqrt{2}}{\sqrt{2}}\bold{i} = \{-30\bold{i} + 50\bold{j}\}\mathrm{~km/h}$$

So that,

$$v_w=\sqrt{(-30{)}^2+(50{)}^2}=\boxed{58.31km/h}$$

Noting that $\bold{v}_w$ has $-\bold{i}$ and $+\bold{j}$ components, as shown in the Fig., its direction is therefore

Answer