Question

A plane flies in a direction of 24.5 degrees south of west at 290 mph. it encounters a 30.5 mph wind that is heading 21 degrees east of north.

Answer 1

The resultant velocity of the plane is 275.5 mph at a direction of 60.7 degrees south of west.

How to solve

1. Convert wind velocity to components:

Northward component
`(V_n) = 30.5 mph * \cos(21 degrees) = 28.6 mph`

Eastward component
`(V_e) = 30.5 mph * \sin(21 degrees) = 10.4 mph`

2. Convert plane velocity to components:

Westward component
`(V_w) = 290 mph * \sin(24.5 degrees) = 125.9 mph`

Southward component
`(V_s) = 290 mph * \cos(24.5 degrees) = 260.1 mph`

3. Combine wind and plane components:

Resultant eastward velocity
`(V_re) = V_e + V_w = 10.4 mph + 125.9 mph = 136.3 mph`

Resultant northward velocity (V_rn) = V_n - V_s = 28.6 mph - 260.1 mph = -231.5 mph (negative indicates southward direction)

4. Calculate resultant magnitude:

Resultant velocity (V_r) =
`\sqrt(V_re^2 + V_rn^2) = sqrt(136.3^2 + (-231.5)^2) = 275.5 mph`

5. Calculate resultant direction:

Inverse tangent of (V_rn / V_re) =
`tan^((-1))(-231.5 / 136.3) = -60.7 degrees`

Therefore, the resultant velocity of the plane is 275.5 mph at a direction of 60.7 degrees south of west.

The Complete Question

A plane flies in a direction of 24.5 degrees south of west at 290 mph. It encounters a 30.5 mph wind that is heading 21 degrees east of north. What is the resultant velocity of the plane (magnitude and direction)?