Questions: Two airplanes leave an airport at the same time. The velocity of the first airplane is 740 m / h at a heading of 39.3°. The velocity of the second is 620 m / h at a heading of 109°. How far apart are they after 2.1 h? Answer in units of m.

Two airplanes leave an airport at the same time. The velocity of the first airplane is 740 m / h at a heading of 39.3°. The velocity of the second is 620 m / h at a heading of 109°. How far apart are they after 2.1 h? Answer in units of m.

Transcript text: Two airplanes leave an airport at the same time. The velocity of the first airplane is $740 \mathrm{~m} / \mathrm{h}$ at a heading of $39.3^{\circ}$. The velocity of the second is $620 \mathrm{~m} / \mathrm{h}$ at a heading of $109^{\circ}$. How far apart are they after 2.1 h ? Answer in units of m .

Solution

Solution Steps

Step 1: Convert Velocities to Components

To find the distance between the two airplanes, we first need to convert their velocities into vector components. The velocity of each airplane can be broken down into horizontal (x-axis) and vertical (y-axis) components using trigonometry.

For the first airplane:

Velocity: $740 \, \text{m/h}$
Heading: $39.3^\circ$

The components are: \[ v_{1x} = 740 \cos(39.3^\circ) \] \[ v_{1y} = 740 \sin(39.3^\circ) \]

For the second airplane:

Velocity: $620 \, \text{m/h}$
Heading: $109^\circ$

The components are: \[ v_{2x} = 620 \cos(109^\circ) \] \[ v_{2y} = 620 \sin(109^\circ) \]

Step 2: Calculate the Position After 2.1 Hours

Next, we calculate the position of each airplane after 2.1 hours using the velocity components.

For the first airplane: \[ x_1 = v_{1x} \times 2.1 \] \[ y_1 = v_{1y} \times 2.1 \]

For the second airplane: \[ x_2 = v_{2x} \times 2.1 \] \[ y_2 = v_{2y} \times 2.1 \]

Step 3: Calculate the Distance Between the Two Airplanes

The distance between the two airplanes can be found using the distance formula: \[ d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \]

Step 4: Perform the Calculations

First, calculate the components:

For the first airplane: \[ v_{1x} = 740 \cos(39.3^\circ) \approx 573.6763 \, \text{m/h} \] \[ v_{1y} = 740 \sin(39.3^\circ) \approx 468.0713 \, \text{m/h} \]

For the second airplane: \[ v_{2x} = 620 \cos(109^\circ) \approx -204.9610 \, \text{m/h} \] \[ v_{2y} = 620 \sin(109^\circ) \approx 585.1945 \, \text{m/h} \]

Now, calculate the positions after 2.1 hours:

For the first airplane: \[ x_1 = 573.6763 \times 2.1 \approx 1204.7202 \, \text{m} \] \[ y_1 = 468.0713 \times 2.1 \approx 982.9497 \, \text{m} \]

For the second airplane: \[ x_2 = -204.9610 \times 2.1 \approx -430.4181 \, \text{m} \] \[ y_2 = 585.1945 \times 2.1 \approx 1228.9085 \, \text{m} \]

Finally, calculate the distance: \[ d = \sqrt{(-430.4181 - 1204.7202)^2 + (1228.9085 - 982.9497)^2} \] \[ d = \sqrt{(-1635.1383)^2 + (245.9588)^2} \] \[ d = \sqrt{2674970.5 + 60405.5} \] \[ d = \sqrt{2735376.0} \approx 1653.8945 \, \text{m} \]