physics:-Velocity,speed

Question

Two airplanes leave an airport at the same

time. The velocity of the first airplane is

660 m/h at a heading of 55.2

◦

. The velocity

of the second is 590 m/h at a heading of 111◦

.

How far apart are they after 2.1 h?

Answer in units of m.

Answer 1

With N as +ve Y direction, and E as +ve X direction:

Velocity of 1st air plane, V_A=660*cos(55.2 degree) j + 660 sin(55.2degree) i mph

Using displacement = velocity * time, find displacement of A from airport,

Velocity of 2nd air plane, V_B=590*cos(111 degree) j + 590 sin(111 degree) i mph

Similarly find displacement of B.

Note that with airport as the origin, displacement is same as the position vector the airplance.

Heading angle is the angle with respect to the north direction. So cosine of the heading angle gives the north component, and the sine the east component.

r_A = Position vector of A = 660*2.1(cos(55.2 degree) j + sin(55.2degree)i) miles

=791 j + 1138 i miles

r_B=Position vector of B = 590*2.1* (cos(111 degree) j + sin(111 degree)i) miles.

=-444 j + 1157 i miles

DIsptance between A and B = magnititude of (r_A- r_B)

Other way:

The angle between two displacement vectors is 111-55.2 = 55.8 degree

Magnitude of vectors are 1386 miles and 1239 miles.

Use cosine law, find the distance as the length of third side of the triangle:

sqrt( 1386^2 + 1239^2 -2*1239*1386cos(55.8degree))