Question

1. a pilot is required to fly directly from london, united kingdom, to rome, italy, in 3.4 h. the displacement is 1.4×10³ km s 43° e. the wind velocity reported from the ground is 55 km/h s. determine the required velocity of the plane relative to the air.

Explanation:

Step1: Calculate the average velocity of the plane relative to the ground

The average velocity of the plane relative to the ground $v_{pg}$ is given by the displacement $\vec{d}$ divided by the time $t$. The magnitude of the displacement $d = 1.4\times10^{3}\text{ km}$ and $t = 3.4\text{ h}$. So, $v_{pg}=\frac{d}{t}=\frac{1.4\times 10^{3}\text{ km}}{3.4\text{ h}}\approx411.76\text{ km/h}$.
The direction of $\vec{v}_{pg}$ is $\text{S }43^{\circ}\text{ E}$.

Step2: Resolve the velocity - vectors into components

Let the velocity of the plane relative to the air be $\vec{v}_{pa}$, the velocity of the wind relative to the ground be $\vec{v}_{wg}=55\text{ km/h [S]}$, and the velocity of the plane relative to the ground be $\vec{v}_{pg}$.
We know that $\vec{v}_{pg}=\vec{v}_{pa}+\vec{v}_{wg}$, so $\vec{v}_{pa}=\vec{v}_{pg}-\vec{v}_{wg}$.
In the x - direction (east - west), $v_{pgx}=v_{pg}\sin43^{\circ}\approx411.76\times\sin43^{\circ}\approx280.7\text{ km/h}$ (east)
In the y - direction (north - south), $v_{pgy}=-v_{pg}\cos43^{\circ}\approx - 411.76\times\cos43^{\circ}\approx - 299.4\text{ km/h}$ (south)
The y - component of $\vec{v}_{wg}$ is $v_{wg y}=- 55\text{ km/h}$ (south)
The y - component of $\vec{v}_{pa}$ is $v_{pay}=v_{pgy}-v_{wg y}=-299.4 - (-55)=-244.4\text{ km/h}$ (south)
The x - component of $\vec{v}_{pa}$ is $v_{pax}=v_{pgx}=280.7\text{ km/h}$ (east)

Step3: Calculate the magnitude of $\vec{v}_{pa}$

The magnitude of $\vec{v}_{pa}$ is given by $v_{pa}=\sqrt{v_{pax}^{2}+v_{pay}^{2}}=\sqrt{(280.7)^{2}+(-244.4)^{2}}\approx372\text{ km/h}$

Step4: Calculate the direction of $\vec{v}_{pa}$

The direction $\theta$ of $\vec{v}_{pa}$ is $\tan\theta=\frac{v_{pax}}{|v_{pay}|}=\frac{280.7}{244.4}\approx1.148$
$\theta=\arctan(1.148)\approx49^{\circ}$ east of south

Answer:

The magnitude of the required velocity of the plane relative to the air is approximately $372\text{ km/h}$ and the direction is $\text{S }49^{\circ}\text{ E}$