Question

A small plane is flying northeast at a speed of 200 mph. A wind blows due west at 28 mph. Find the resultant speed and direction of the plane.

Answer 1

The resultant speed = 181.3 mph

The final direction = 38.7° northeast.

Explanation:

We need to find the component in the x-direction and in the y-direction of the speed:

For the plane:

`v_{p_(x)} = 200cos(45) = 141.42`

`v_{p_(y)} = 200sin(45) = 141.42`

For the wind we have:

`v_{w_(x)} = -28`

`v_{w_(y)} = 0`

Now, the total speed in the x-direction and in the y-direction is:

`V_(x) = v_{p_(x)} + v_{w_(x)} = 141.42 - 28 = 113.42`

`V_(y) = v_{p_(y)} + v_{w_(y)} = 141.42`

Hence, the resultant speed is:

`V = \sqrt{V_(x)^(2) + V_(y)^(2)} = \sqrt{(113.42)^(2) + (141.42)^(2)} = 181.3 mph`

Finally, the direction of the plane is:

`tan(\theta) = (V_(y))/(V_(x)) = (141.42)/(113.42) = 1.25`

`\theta = 51.3 ^(\circ)`

`\theta = 90 - 51.3 = 38.7 ^(\circ)`

The plane is moving at 38.7° northeast.

I hope it helps you!