# Airplane. What is the velocity?

What is the velocity at which an airplane must fly in order to cover the distance of 400 km to the north in 1 hour? If the wind is blowing to the north east (30° east of north) at 140 km/hr.

###### Solution. Posted on Friday, November 21, 2014.

The relative velocity law for two dimensions (non-relativistic) said that velocity of the object A relative to the non-moving object equal to the vector sum of the velocity of the object A relative to the object B and the velocity of the object B relative to the non-moving object.

V(airplane-Earth) = V(airplane-wind) + V(wind-Earth)

In order to solve this problem, let’s use the law of cosines .

c2 = a2 + b2 − 2ab cos(C)

V(airplane-wind)2 = 4002 + 1402 − 2*400*140*cos(30)

V(airplane-wind) = 287.41 km/hr

In order to find direction, let’s use the law of cosines again.

1402 = 287.412 + 4002 − 2*287.41*400*cos(C)

C= 14.10 West of North.

## 2 thoughts on “Airplane. What is the velocity?”

1. Johnny B says:

The plane goes North (N) 400 km in 1 hr.

The 400 km/hr velocity is made
up of the N-comp of the wind plus the
N-comp of velocity of plane.

The N-comp of the wind is
140 km/hr * cos(30) = 121.24 km/hr.

The N-comp of the plane vel is
400 km/hr – 121.24 km/hr = 278.76 km/hr.

The magnitude of the W-comp of the
plane velocity is = magnitude of the
E-comp of the wind velocity.
The E-comp of the plane’s velocity
must balance the W-comp of the wind
for the plane to go directly North.
So,
magnitude W-comp of plane vel =
magnitude E-comp of wind =
140 km/hr * sin(30) = 70 km/hr.

One way to find the mag of plane’s velocity
is to apply the Pythagorean Theorem.

sqrt ( (278.76)^2 + (70)^2 ) = 287.41 km/hr

The direction the plane must fly
can be found by finding the
arctan (70/278.76) = 14.1 deg W of N.

Vel of plane = 287.41 km/hr at 14.1 deg W of N.

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