Question

A jet airliner, moving initially at $300mi/h$ to the east, suddenly enters a region where the wind is blowing at$100mi/h$ towards the direction $30.0^{\circ }$north of east. What are the new speed and direction of the aircraft relative to the ground?

Short answer

The new speed and direction of an aircraft is $389.82km/h$ towards the direction$7.{37}^{\circ }$ north to east.

Step-by-step solution

Step 1:finding the new speed of an aircraft

Representing the motion of aircraft as $\stackrel{\perp }{a}$,the wind as $\stackrel{\perp }{b}$,speed and direction of the aircraft be $\stackrel{\perp }{c}$,then

$\stackrel{\perp }{c}=\stackrel{\perp }{a}+\stackrel{\perp }{b}$

Finding the projections of vectors on x and y axis.

$\stackrel{\perp }{a}$ along x and y axis are, , . localid="1663829188309" $\text{x}\stackrel{\perp }{a}=300,y\stackrel{\perp }{a}=0$

$\stackrel{\perp }{b}$along x and y axis are,, $$\begin{gathered} \text{x}\stackrel{r}{b}=100\left(cos{30}^{\circ }\right)=100\frac{\sqrt{3}}{2}=50\sqrt{3} \\ y\stackrel{\perp }{b}=100\left(sin{30}^{\circ }\right)=50 \\ x\stackrel{\perp }{c}=x\stackrel{\perp }{a}+x\stackrel{\perp }{b}=300+50\sqrt{3}=386.60 \end{gathered}$$

Therefore, $y\stackrel{r}{c}=y\stackrel{r}{a}+y\stackrel{r}{b}=0+50=50$

$\left|\stackrel{r}{c}\right|=\sqrt{x\stackrel{r}{c^2}+y\stackrel{r}{c^2}}=\sqrt{386.{60}^2+{50}^2}=389.82km/h$

Finding the direction of an aircraft

$y=arc\sin \left(\frac{y{\displaystyle \stackrel{\perp }{c}}}{{\displaystyle \stackrel{r}{\left|c\right|}}}\right)=arc\sin \left(\frac{50}{389.82}\right)=arc\sin (0.1283)$

To find the direction,

=$7.{37}^{\circ }$

Therefore ,the new speed and direction of an aircraft is $\mathit{389}\mathit{.}\mathit{82}km\mathit{/}h$ towards$7.{37}^{\circ }$ north to east.