Chapter 3, Problem 58GP

To determine

The velocity of wind

Answer to Problem 58GP

The velocity of wind is 110.6 km/hr to the 37 degree north of east.

Explanation of Solution

Given:

  $\begin{array}{l}\text{Velocity of plane, }{v}_p=155\text{ km/hr}\\ \text{Velocity of plane after 1 hour, }{v}_1=125\text{ km/hr}\\ \text{Velocity of wind, }{v}_w\end{array}$

Calculation:

The vector triangle is drawn as:

  

Where,

  $\begin{array}{l}{V}_{WG}=\text{velocity of wind with respect to ground}\\ V_{PW}=\text{velocity of plane with respect to wind}\\ V_{PG}=\text{velocity of plane with respect to ground}\end{array}$

From figure:

  $V_{PG}=V_{PW}+V_{WG}$

All are in vector form.

Now, the x component of the velocities are given as:

  $\begin{array}{l}{V}_{W{G}_x}=V_{P{G}_x}\\ =V_{PG}\cos {45}^o\\ =125\left(0.7071\right)\\ =88.4\text{km/h}\end{array}$

Similarly, the y component:

  $\begin{array}{l}{V}_{W{G}_y}=V_{P{G}_y}-V_{PW}\\ =-V_{PG}\sin {45}^o-\left(-155\right)\\ =125\sin {45}^o+155\\ =66.6\text{km/h}\end{array}$

Now, the resultant of the x and the y component will be the magnitude of the wind velocity:

  $\begin{array}{l}\left|V_{WG}\right|=\sqrt{{\left(V_{W{G}_X}\right)}^2+{\left(V_{W{G}_y}\right)}^2}\\ =110.6\text{km/h}\end{array}$

Now, evaluating the angle:

  $\begin{array}{l}\tan \theta =\frac{V_{W{G}_N}}{V_{W{G}_E}}\\ \theta ={\tan }^{-1}\left(\frac{66.6}{88.4}\right)\\ ={37}^o\end{array}$

The direction is 37 degree north east.