Question

An airplane is flying on a bearing of 330 degrees at 450 mph. Find the component form of the velocity of the airplane.

Answer 1

The component form of the velocity of the airplane is
`\vec v = 389.711\,\hat{i} -225\,\hat{j}\,\left[(m)/(s) \right]`.

Explanation:

Let suppose that a bearing of 0 degrees corresponds with the
`+x` direction and that angle is measured counterclockwise. Besides, we must know both the magnitude of velocity (
`\|\vec v\|`), in miles per hour, and the direction of the airplane (
`\theta`), in sexagesimal degrees to construct the respective vector. The component form of the velocity of the airplane is equivalent to a vector in rectangular form with physical units, that is:

`\vec v = \|\vec v\|\cdot (\cos \theta \,\hat{i}+\sin \theta\,\hat{j})` (1)

If we know that
`\|\vec v\| = 450\,(mi)/(h)` and
`\theta = 330^(\circ)`, then the component form of the velocity of the airplane is:

`\vec v = 389.711\,\hat{i} -225\,\hat{j}\,\left[(m)/(s) \right]`

The component form of the velocity of the airplane is
`\vec v = 389.711\,\hat{i} -225\,\hat{j}\,\left[(m)/(s) \right]`.