49

Q 49. An aeroplane has to go from a point A to another point B, 500 km away due 30° east of north. A wind is blowing due north at a speed of 20 m/s. The air-speed of the plane is 150 m/s. (a) Find the direction in which the pilot should head the plane to reach the point B. (b) Find the time taken by the plane to go from A to B.

I will form the diagram for the given problem.

Let us first assign the unknown variables.
t = Total time of flight (unknown)
A = Angle of destination vector from north = 30∘
B = Angle of airplane's propulsion, measured from East (to find)
V = Final Velocity (Unknown)
u = Air Velocity = 20ms−1
s = Airplane Velocity = 150ms−1

The air velocity only provides a northward displacement, ie the Eastward displacement should come from the East component of the plane's velocity.

The eastward displacement is 500×cos(90∘−A)km
The East component of the airplane's velocity is s×cos(B)

So the equation is 
500×cos(90∘−A)=s×cos(B)×t

You will obtain an equation in B and t.

Now, the northward displacement is due to both the airplane's propulsion and the wind velocity.
The net northward component of velocity is
s×sin(B)+u

The distance is 
500×sin(90∘−A)

Thus the equation is
500×sin(90∘−A)=(s×sin(B)+u)×t

You have another equation in B and t.

Solve for B