Question #136674

A jet airliner, moving initially at 300 Km/h to the east, suddenly enters a region where the wind is blowing at 100 Km/h toward the direction 30.00 north of east. What is the new speed and direction of the aircraft relative to the ground?

Expert's answer

Solution.

The task is shown in the following picture:



If we represent the motion of the aircraft as a vector a⃗\vec a , the wind as a vector b⃗\vec b and the new speed and direction of the aircraft as a vector c⃗\vec c then:

c⃗=a⃗+b⃗\vec c=\vec a+\vec b .


Find projections of the vectors on the axis x and y:

xa⃗=300;x_{\vec a}=300; ya⃗=0.y_{\vec a}=0.

xb⃗=100⋅cos(30°)=10032=503;x_{\vec b}=100\cdot cos(30\degree)=100\frac{\sqrt{3}}{2}=50\sqrt{3}; yb⃗=100⋅sin(30°)=100⋅12=50.y_{\vec b}=100\cdot sin(30\degree)=100\cdot\frac{1}{2}=50.

and xc⃗=xa⃗+xb⃗=300+503≈386.60;x_{\vec c}=x_{\vec a}+x_{\vec b}=300+50\sqrt{3}\approx386.60; yc⃗=ya⃗+yb⃗=0+50=50.y_{\vec c}=y_{\vec a}+y_{\vec b}=0+50=50.


Find the length of the vector c⃗\vec c :

∣c⃗∣=xc⃗2+yc⃗2≈386.602+502≈389.82|\vec c|=\sqrt{x_{\vec c}^2+y_{\vec c}^2}\approx\sqrt{386.60^2+50^2}\approx389.82 Km/h.

Find the angle γ:

γ=arcsin(yc⃗∣c⃗∣)≈arcsin(50389.82)≈arcsin(0.1283)≈7.37°γ=arcsin(\frac{y_{\vec c}}{|\vec c|})\approx arcsin(\frac{50}{389.82})\approx arcsin(0.1283)\approx7.37\degree .


Answer: the new speed and direction of the aircraft is 389.82 Km/h toward the direction 7.37° north of east.


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