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?
1
Expert's answer
2020-10-13T07:12:42-0400

Solution.

The task is shown in the following picture:



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

c=a+b\vec c=\vec a+\vec b .


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

xa=300;x_{\vec a}=300; ya=0.y_{\vec a}=0.

xb=100cos(30°)=10032=503;x_{\vec b}=100\cdot cos(30\degree)=100\frac{\sqrt{3}}{2}=50\sqrt{3}; yb=100sin(30°)=10012=50.y_{\vec b}=100\cdot sin(30\degree)=100\cdot\frac{1}{2}=50.

and xc=xa+xb=300+503386.60;x_{\vec c}=x_{\vec a}+x_{\vec b}=300+50\sqrt{3}\approx386.60; yc=ya+yb=0+50=50.y_{\vec c}=y_{\vec a}+y_{\vec b}=0+50=50.


Find the length of the vector c\vec c :

c=xc2+yc2386.602+502389.82|\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(ycc)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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