Question

An airplane is heading due south at a speed of 688 km/h. If a wind begins blowing from the southwest at a speed of 90.0 km/h (average), calculate (a) the velocity (magnitude and direction) of the plane relative to the ground and (b) how far from its intended position it will be after 11.0 min if the pilot takes no corrective action. [Hint: First, draw a diagram.]

Short answer

(a) The velocity of the plane relative to the ground is$628\mathrm{km}/\mathrm{h},$and the direction is $5.81°$east of south.

(b) The distance moved by the plane due to the air from its intended position is $16.5\mathrm{km}$.

Step-by-step solution

Step 1. Meaning of relative motion

An object's relative motion may be defined as the movement of the object in relation to a particular frame of reference. It depends on the point of view or frame of reference.

Step 2. Draw the vector diagram

Here, ${\overrightarrow{v}}_{ag}$is the velocity of the air with respect to the ground, ${\overrightarrow{v}}_{pa}$is the velocity of the plane with respect to the air, and ${\overrightarrow{v}}_{pg}$is the velocity of the plane with respect to the ground. Here, $\theta$is the angle made by ${\overrightarrow{v}}_{pg}$. with the vertical. The velocity ${\overrightarrow{v}}_{ag}$makes an angle of $45°$with the horizontal.

Step 3. Calculate the velocity of the plane with respect to the air

(a)

The velocity of the plane with respect to the air can be calculated as

$\begin{array}{c}{\overrightarrow{v}}_{pa}=\left(v_{pa}\cos \left(-90°\right)\right)\hat{i}+\left(v_{pa}\sin \left(-90°\right)\right)\hat{j}\\ {\overrightarrow{v}}_{pa}=\left(0\right)\hat{i}+\left(v_{pa}\sin \left(-90°\right)\right)\hat{j}\\ {\overrightarrow{v}}_{pa}=\left(v_{pa}\sin \left(-90°\right)\right)\hat{j}\end{array}$

Step 4. Calculate the velocity of the air with respect to the ground

The velocity of the air with respect to the ground can be calculated as

${\overrightarrow{v}}_{ag}=\left(v_{ag}\cos 45°\right)\hat{i}+\left(v_{ag}\sin 45°\right)\hat{j}$.

Step 5. Calculate the velocity of the plane with respect to the ground

Thevelocity of the plane with respect to the ground can be calculated as

$\begin{array}{c}{\overrightarrow{v}}_{pg}={\overrightarrow{v}}_{ag}+{\overrightarrow{v}}_{pa}\\ {\overrightarrow{v}}_{pg}=\left(v_{ag}\cos 45°\right)\hat{i}+\left(v_{ag}\sin 45°\right)\hat{j}+\left(v_{pa}\sin \left(-90°\right)\right)\hat{j}\\ {\overrightarrow{v}}_{pg}=\left[\left(90.0\mathrm{km}/\mathrm{h}\right)\cos 45°\right]\hat{i}+\left[\left(90.0\mathrm{km}/\mathrm{h}\right)\sin 45°\right]\hat{j}+\left[\left(688\mathrm{km}/\mathrm{h}\right)\sin \left(-90°\right)\right]\hat{j}\\ {\overrightarrow{v}}_{pg}=\left(63.63\mathrm{km}/\mathrm{h}\right)\hat{i}-\left(624.36\mathrm{km}/\mathrm{h}\right)\hat{j}\end{array}$

Step 6. Calculate the magnitude of the velocity of the plane with respect to the ground

The magnitude of the velocity of the plane with respect to the ground can be calculated as

$\begin{array}{c}\left|{\overrightarrow{v}}_{pg}\right|=\sqrt{v_x^2+v_y^2}\\ \left|{\overrightarrow{v}}_{pg}\right|=\sqrt{{\left(63.63\mathrm{km}/\mathrm{h}\right)}^2+{\left(-624.36\mathrm{km}/\mathrm{h}\right)}^2}\\ \left|{\overrightarrow{v}}_{pg}\right|\approx 628\mathrm{km}/\mathrm{h}\end{array}$

Step 7. Calculate the direction of the velocity of the plane with respect to the ground

The direction of the velocity of the plane with respect to the ground can be calculated as

$\begin{array}{l}\theta ={\tan }^{-1}\left(\frac{v_x}{v_y}\right)\\ \theta ={\tan }^{-1}\left(\frac{63.63\mathrm{km}/\mathrm{h}}{-624.36\mathrm{km}/\mathrm{h}}\right)\\ \theta =-5.81°\end{array}$

Thus, the velocity of the plane relative to the ground is$628\mathrm{km}/\mathrm{h}$, and the direction is $5.81°$east of south.

Step 8. Calculate the distance of the plane from the intended position

(b)

The distance of the plane from the intended position can be calculated as

$\begin{array}{l}\Delta x=v_{ag}t\\ \Delta x=\left(90.0\mathrm{km}/\mathrm{h}\right)\left(11\min \right)\left(\frac{1\mathrm{h}}{60\min }\right)\\ \Delta x=16.5\mathrm{km}\end{array}$

Thus, the distance moved by the plane due to the air from its intended position is $16.5\mathrm{km}$.