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 .

c^{2} = a^{2} + b^{2} − 2ab cos(C)

V(airplane-wind)^{2} = 400^{2} + 140^{2} − 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.

140^{2} = 287.41^{2} + 400^{2} − 2*287.41*400*cos(C)

C= 14.1^{0} West of North.

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.

LikeLiked by 1 person

Thank you for your solution !

LikeLike