Question

A jet moving at 500.0 km/ h due is in a region where the wind is moving at 120.0 km/h in a direction 30.00° north of east. What is the speed of the aircraft relative to the ground?

Answer 1

The speed of the aircraft relative to the ground is 606.9
`m/s^2`

Step-by-step explanation:

x-y coordinate system:

x is Positive due East direction. Similarly y is Positive due North Direction.

Now let us Decompose each vector into x-y

500 km/h due east = (500, 0)

120 km/h at 30 degree north of east

=
`(120 cos(30), 120 * sin(30))`

=
`(120 * (√((3)))/(2), 120 * (1)/( 2))`

=
`(60 * \sqrt(3), 60)`

Adding the vectors.

=
`(500, 0) + (60 * √(3), 60)`

=
`(500 + 60 * \sqrt(3), 60)`

=
`(500 + 60 * 1.73, 60)`

=
`(500 +103.8, 60)`

= (603.8, 60)

Returning back to polar form

Magnitude =
`√(603.923^2 + 60^2) = 606.9`