Question #120399
An airplane pilot wishes to fly due West. A wind of 80.0 km/h is blowing towards the South.(a) if the airspeed of the plane (its speed in still air) is 320.0 km/h in which direction should the pilot head ? (B) what is the speed of the plane over the ground? draw a vector diagram.
1
Expert's answer
2020-06-08T19:10:36-0400

Let

vp/g=\overrightarrow{v}_{p/g}= the velocity of the plane relative to the ground (due west),

vp/a=\overrightarrow{v}_{p/a}= the velocity of the plane relative to the air (vp/a=320.0 km/h|\overrightarrow{v}_{p/a}|=320.0\ km/h ), and

va/g=\overrightarrow{v}_{a/g}= the velocity of the air relative to the ground (va/g=80.0 km/h,|\overrightarrow{v}_{a/g}|=80.0\ km/h, due south).



The velocity vector addition


vp/a+va/g=vp/g\overrightarrow{v}_{p/a}+\overrightarrow{v}_{a/g}=\overrightarrow{v}_{p/g}

a)

sinθ=va/gvp/a=40km/h320km/h=0.25\sin \theta={|\overrightarrow{v}_{a/g}|\over |\overrightarrow{v}_{p/a}|}={40 km/h\over 320 km/h}=0.25

θ=14.5°,north of west\theta=14.5\degree,\text{north of west}

b)


vp/q=(vp/a)2(va/g)2=|\overrightarrow{v}_{p/q}|=\sqrt{(|\overrightarrow{v}_{p/a}|)^2-(|\overrightarrow{v}_{a/g}|)^2}=

=(320 km/h)2(80 km/h)2=8015 km/h=\sqrt{(320 \ km/h)^2-(80\ km/h)^2}=80\sqrt{15}\ km/h\approx

310 km/h\approx310\ km/h


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