Chapter 3, Problem 49P

To determine

The direction in which the plane head.

Answer to Problem 49P

  $27.16°\text{ south of west}$

Explanation of Solution

Given:

  $\begin{array}{l}\text{Speed of plane, }{v}_p=620\text{ km/hr}\\ \text{Angle, }\theta \text{ = 35°}\\ \text{Speed of wind, }{v}_w=95\text{ km/hr}\end{array}$

Formula Used:

Relative velocity

  $v_{pw}=v_p-v_w$

Calculation:

Consider the given situation.

  

  $\begin{array}{l}{v}_{pw}=v_p\cos 35°-\left[v_p\sin 35°-v_w\right]\\ v_{pw}=620\cos 35°-\left[620\sin 35°-95\right]\\ v_{pw}=507.87i-\left[260.61\right]j\\ v_{pw}=507.87i-260.61j\end{array}$

Now,

  

  $\begin{array}{l}\tan \theta =\frac{260.61}{207.81}\\ \theta =27.16°\text{ south of west}\end{array}$

Conclusion:

Therefore, the direction is $27.16°\text{ south of west}$