In a 2D game environment an airplane trip involves 3 legs, with two stopovers. The first leg is due east for 620km, the second leg is southeast for 440km and the third leg is at 53° of west, for 550km. 
What is the plane's total displacement vector by magnitude (km) and angle (°)?

To find the total displacement vector of the airplane trip, we need to break each leg of the trip into its component vectors and then sum them.

### Step 1: Break Down Each Leg into Components

1. **First Leg (620 km due East)**
   - \( x_1 = 620 \, \text{km} \)
   - \( y_1 = 0 \)

2. **Second Leg (440 km Southeast)**
   - Southeast means \( 45^\circ \) south of east.
   - \( x_2 = 440 \cos(45^\circ) = 440 \times \frac{\sqrt{2}}{2} \approx 311.77 \, \text{km} \)
   - \( y_2 = -440 \sin(45^\circ) = -440 \times \frac{\sqrt{2}}{2} \approx -311.77 \, \text{km} \)

3. **Third Leg (550 km at 53° West of North)**
   - This angle is measured from the north direction, so we convert it to the standard position angle:
   - \( \theta = 90^\circ + 53^\circ = 143^\circ \)
   - \( x_3 = 550 \cos(143^\circ) \approx 550 \times -0.8192 \approx -450.56 \, \text{km} \)
   - \( y_3 = 550 \sin(143^\circ) \approx 550 \times 0.5736 \approx 315.48 \, \text{km} \)

### Step 2: Sum the Components

Now, sum the x and y components.

- **Total x-component:**
  \[
  x_{\text{total}} = x_1 + x_2 + x_3 = 620 + 311.77 - 450.56 \approx 481.21 \, \text{km}
  \]

- **Total y-component:**
  \[
  y_{\text{total}} = y_1 + y_2 + y_3 = 0 - 311.77 + 315.48 \approx 3.71 \, \text{km}
  \]

### Step 3: Calculate the Magnitude and Angle of the Resultant Vector

1. **Magnitude:**
   \[
   R = \sqrt{x_{\text{total}}^2 + y_{\text{total}}^2} = \sqrt{(481.21)^2 + (3.71)^2} \approx \sqrt{231,013.21 + 13.76} \approx \sqrt{231,027} \approx 481.21 \, \text{km}
   \]

2. **Angle:**
   \[
   \theta = \tan^{-1}\left(\frac{y_{\text{total}}}{x_{\text{total}}}\right) = \tan^{-1}\left(\frac{3.71}{481.21}\right) \approx 0.46^\circ
   \]

### Final Results

- **Total Displacement Magnitude:** Approximately **481.21 km**
- **Total Displacement Angle:** Approximately **0.46°** (measured from the east direction)

If you have any questions or need further details, feel free to ask!

### Related Questions
1. How would the displacement change if the second leg were instead due south?
2. What is the average speed if the trip took 2 hours?
3. How would you calculate the total time taken for each leg if you know the speed?
4. What would be the displacement if the third leg were directed due west instead?
5. How would the total displacement differ in a 3D environment? 

**Tip:** Always break down vector components to simplify calculations of displacement and resultant vectors.

In a 2D game environment an airplane trip involves 3 legs, with two stopovers. The first leg is due east for 620km, the second leg is southeast for 440km and the third leg is at 53° of west, for 550km. What is the plane's total displacement vector by magnitude (km) and angle (°)?

Discover how to calculate the total displacement of an airplane trip involving three legs: 620 km east, 440 km southeast, and 550 km at 53° west. The total displacement vector magnitude is approximately 481.21 km, with an angle of about 0.46° from the east. Ideal for high school students learning about vectors.

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