Question

An airplane is taking off headed due north with an air speed of 173 miles per hour at an angle of 18° relative to the

horizontal. The wind is blowing with a velocity of 42 miles per hour at an angle of S47°E. Find a vector that represents the
resultant velocity of the plane relative to the point of takeoff. Let i point east, j point north, and k point up.

Answer 1

`v_\text{plane} = 30.72\; \vec{i} + 193.18\; \vec{j} + 53.46\; \vec{k}`.

Explanation:

Refer to the first diagram:

• Component of the plane's velocity in the direction of vector
  `\vec{j}` relative to the air:
  `173 \cos{18^(\circ)}`;
• Component of the plane's velocity in the direction of vector
  `\vec{k}` relative to the air:
  `173\sin{18^(\circ)}`.

The direction of the plane's velocity relative to the air is normal to vector
`\vec{i}`. Therefore, the component of the plane's velocity (relative to the air) in that direction will equal zero. Thus

`\vec{v}_\text{velocity of plane relative to air} = 0 \; \vec{i} + (173\;\cos{18^(\circ)})\; \vec{j} + (173\;\sin{18^(\circ)})\; \vec{k}`.

Refer to the second diagram,

• Component of the velocity of the wind in the direction of vector
  `\vec{i}`:
  `42 \sin{47^(\circ)}`;
• Component of the velocity of the wind in the direction of vector
  `\vec{j}`:
  `- 42 \cos{47^(\circ)}`.

Assume that the wind blows horizontally. The direction of the wind will be normal to vector
`\vec{k}`. The component of the velocity of the wind in the direction of vector
`\vec{k}` will thus equal zero. Therefore,

`\vec{v}_\text{wind} = (42\sin{47^(\circ)})\;\vec{i} + (42\cos{47^(\circ)})\;\vec{j} + 0\;\vec{k}`.

The ground speed of the plane
`\vec{v}_\text{velocity of plane relative to ground}` is the sum of
`\vec{v}_\text{velocity of plane relative to air}` and
`\vec{v}_\text{wind}`.

That is:

`\begin{aligned}&\vec{v}_\text{velocity of plane relative to ground} \\ =& \vec{v}_\text{velocity of plane relative to air} +\vec{v}_\text{wind} \\ =& \left[0 \; \vec{i} + (173\;\cos{18^(\circ)})\; \vec{j} + (173\;\sin{18^(\circ)})\; \vec{k}\right] + \\ &\left[(42\sin{47^(\circ)})\;\vec{i} + (42\cos{47^(\circ)})\;\vec{j} + 0\;\vec{k} \right] \\= &  30.72\; \vec{i} + 193.18\; \vec{j} + 53.46\; \vec{k} \end{aligned}`.