Chapter 6.3, Problem 37SSC

To determine

Plane speed and direction relative to ground.

Answer to Problem 37SSC

Speed of plane relative to ground is $v_{pg}=194.56\text{km/h}$ and direction is $\theta =64.09°$ south east.

Explanation of Solution

Given:

Speed of plane relative to air is $v_{pa}=175\text{km/h}$ towards south

Speed of wind relative to ground is $v_{ag}=85\text{km/h}$ towards east

Formula used:

The resultant of two perpendicular vectors is

  $R=\sqrt{x^2+y^2}$

Angle of resultant is

  $\theta ={\tan }^{-1}\frac{x}{y}$

Calculation:

Speed of plane relative to ground is

  $\begin{array}{c}{v}_{pg}{}^2=v_{pa}{}^2+v_{ag}{}^2\\ v_{pg}=\sqrt{v_{pa}{}^2+v_{ag}{}^2}\\ =\sqrt{{175}^2+{85}^2}\\ =194.56\text{km/h}\end{array}$

Direction of plane relative to ground is,

  $\begin{array}{c}\theta ={\tan }^{-1}\left(\frac{v_{pa}}{v_{ag}}\right)\\ ={\tan }^{-1}\left(\frac{175}{85}\right)\\ \theta =64.09°\end{array}$

So, direction is $\theta =64.09°$ south east.

Conclusion:

Therefore, speed of plane relative to ground is $v_{pg}=194.56\text{km/h}$ and direction is $\theta =64.09°$ south east.