Question

Two airplanes leave an airport at the same time. The velocity of the first airplane is 730 m/h at a heading of 53.6 degrees. The velocity of the second is 570 m/h at a heading of 88 degrees. How far apart are they after 3.3 h? Answer in units of m.

Answer 1

To find how far apart the two airplanes are after 3.3 hours, we will analyze their respective positions using vector components based on their velocities and headings.

### Step 1: Determine the velocities in component form

1. **For the first airplane:**
- Velocity \( V_1 = 730 \) m/h
- Heading = 53.6 degrees

The components of the velocity can be calculated using trigonometric functions:
\[
V_{1x} = V_1 \cdot \cos(53.6^\circ)
\]
\[
V_{1y} = V_1 \cdot \sin(53.6^\circ)
\]

2. **For the second airplane:**
- Velocity \( V_2 = 570 \) m/h
- Heading = 88 degrees

Similarly, we calculate the components:
\[
V_{2x} = V_2 \cdot \cos(88^\circ)
\]
\[
V_{2y} = V_2 \cdot \sin(88^\circ)
\]

### Step 2: Calculate the components

Using a calculator:
1. For the first airplane:
\[
V_{1x} = 730 \cdot \cos(53.6^\circ) \approx 730 \cdot 0.599 = 436.27 \text{ m/h}
\]
\[
V_{1y} = 730 \cdot \sin(53.6^\circ) \approx 730 \cdot 0.800 = 584.00 \text{ m/h}
\]

2. For the second airplane:
\[
V_{2x} = 570 \cdot \cos(88^\circ) \approx 570 \cdot 0.0349 = 19.89 \text{ m/h}
\]
\[
V_{2y} = 570 \cdot \sin(88^\circ) \approx 570 \cdot 0.999 = 569.43 \text{ m/h}
\]

### Step 3: Calculate the positions after 3.3 hours

Next, determine the total displacement for each airplane after \( t = 3.3 \) hours.

1. **For the first airplane:**
\[
x_1 = V_{1x} \cdot t = 436.27 \cdot 3.3 \approx 1440.78 \text{ m}
\]
\[
y_1 = V_{1y} \cdot t = 584.00 \cdot 3.3 \approx 1927.20 \text{ m}
\]

2. **For the second airplane:**
\[
x_2 = V_{2x} \cdot t = 19.89 \cdot 3.3 \approx 65.96 \text{ m}
\]
\[
y_2 = V_{2y} \cdot t = 569.43 \cdot 3.3 \approx 1888.80 \text{ m}
\]

### Step 4: Calculate the distance between the two airplanes

Now, we find the distance \( d \) between the two final positions \( (x_1, y_1) \) and \( (x_2, y_2) \) using the distance formula:
\[
d = \sqrt{(x_1 - x_2)^2 + (y_1 - y_2)^2}
\]

Substituting the calculated values:
\[
d = \sqrt{(1440.78 - 65.96)^2 + (1927.20 - 1888.80)^2}
\]
\[
d = \sqrt{(1374.82)^2 + (38.40)^2}
\]
Calculating the squares:
\[
d = \sqrt{1897910.1284 + 1476.096}
\]
\[
d = \sqrt{1899386.2244} \approx 1379.54 \text{ m}
\]

Thus, the distance between the two airplanes after 3.3 hours is approximately **1379.54 m**.