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

Answer:

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

Step-by-step 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:

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