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.)).

b) What was the velocity (Hint: Treat the plane and wind as having constant speeds. Think about the displacement of the plane—and the time taken— as seen by someone on the ground.)

PIG

c) What was $\vec{v}_{P/A}$ ?

d) What was $\vec{v}_{A/G}$ ?

e) What was the speed ${v}_{A/G}$ ?

f) What was the direction of the wind?

Answer 1

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