A light airplane attains an airspeed (speed of the plane with respect to the air

Question

A light airplane attains an airspeed (speed of the plane with respect to the air) of 500km/hr. The pilot sets out for a destination 800km due north, but discovers that the plane must be pointed 20.0 east of due north for the plane to reach there directly (i.e., the nose/front of the plane is pointed 20.0 east of due north, but the plane itself moves due north as seen by someone on the ground). The plane arrives in 2.00hr. P, A, and G refer to the plane, air and ground, respectively. Answer the following questions, showing formulas and steps as appropriate:

a) Specify your coordinate axes (which way does the X axis point, which way does the Y axis point (north, south, east, west, etc.)).

f) What was the direction of the wind?

Explanation / Answer

part a:

x axis points towards east , y axis points towards north.

part b:

plane’s velocity w.r.t. air=500 km/hr, 20 degree east of north

=500*(sin(20) i+ cos(20) j)

air velocity is let vx i + vy j.

total velocity is along due north i.e. along y axis.

hence x component of plane’s velocity w.r.t. air and velocity of air cancels out.

hence vx+500*sin(20)=0

==>vx=-171 km/hr

total velocity along y axis=distance/time=800/2=400 km/hr

then 500*cos(20)+vy=400

==>vy=-69.85 km/hr

so Vp/g=400 j km/hr

part c: Vp/a=500*(sin(20) i+ cos(20) j)=171 i +469.85 j

part d:

va/g=vx i +vy j

=-171 i-69.85 j km/hr

part e:

speed of Va/g=sqrt(171^2+69.85^2)=184.72 km/hr

part f:

direction of wind=arctan(-69.85/(-171))=22.219 degrees south of west

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