Question: For Questions 14 and 15, use this flight.
An airplane flies three successive straight legs. The table gives its speed and heading relative to the air, held constant on each leg. Neglect time spent turning. East is +x and north is +y.
Leg Airspeed Heading relative to the air Duration
1 150 m/s 45° north of east 20 min
2 180 m/s due east 50 min
3 100 m/s 45° south of east 30 min
The air is still during legs 1 and 2. Only during leg 3, a steady wind blows toward the east at 20 m/s relative to the ground. The pilot keeps the listed airspeed and heading. Questions 14 and 15 refer to the actual flight, including the wind.
14. What is the approximate magnitude of the total ground displacement over all three legs?
(A) 927 km
(B) 831 km
(C) 795 km
(D) 36 km
(E) 915 km
15. Approximately what is the average speed relative to the ground over the entire flight?
(A) 154.5 m/s
(B) 138.4 m/s
(C) 150 m/s
(D) 126 m/s
(E) 170 m/s
1. **State the problem:**
Calculate the total ground displacement magnitude and the average ground speed of an airplane flying three legs with given airspeeds, headings, durations, and wind conditions.
2. **Given data:**
- Leg 1: Airspeed $150\ \mathrm{m/s}$, heading $45^\circ$ north of east, duration $20$ min
- Leg 2: Airspeed $180\ \mathrm{m/s}$, heading due east, duration $50$ min
- Leg 3: Airspeed $100\ \mathrm{m/s}$, heading $45^\circ$ south of east, duration $30$ min
- Wind during leg 3: $20\ \mathrm{m/s}$ eastward
3. **Convert durations to seconds:**
- $20$ min $= 20 \times 60 = 1200\ \mathrm{s}$
- $50$ min $= 50 \times 60 = 3000\ \mathrm{s}$
- $30$ min $= 30 \times 60 = 1800\ \mathrm{s}$
4. **Calculate displacement vectors for each leg relative to ground:**
**Leg 1 (no wind):**
- Velocity components:
$$v_{1x} = 150 \cos 45^\circ = 150 \times \frac{\sqrt{2}}{2} = 106.07\ \mathrm{m/s}$$
$$v_{1y} = 150 \sin 45^\circ = 106.07\ \mathrm{m/s}$$
- Displacement:
$$d_{1x} = v_{1x} \times 1200 = 106.07 \times 1200 = 127284\ \mathrm{m}$$
$$d_{1y} = v_{1y} \times 1200 = 127284\ \mathrm{m}$$
**Leg 2 (no wind):**
- Velocity components:
$$v_{2x} = 180\ \mathrm{m/s}, \quad v_{2y} = 0$$
- Displacement:
$$d_{2x} = 180 \times 3000 = 540000\ \mathrm{m}$$
$$d_{2y} = 0$$
**Leg 3 (with wind):**
- Airspeed components:
$$v_{3x,air} = 100 \cos 45^\circ = 70.71\ \mathrm{m/s}$$
$$v_{3y,air} = -100 \sin 45^\circ = -70.71\ \mathrm{m/s}$$
- Wind velocity:
$$v_{wind,x} = 20\ \mathrm{m/s}, \quad v_{wind,y} = 0$$
- Ground velocity components:
$$v_{3x} = v_{3x,air} + v_{wind,x} = 70.71 + 20 = 90.71\ \mathrm{m/s}$$
$$v_{3y} = v_{3y,air} + 0 = -70.71\ \mathrm{m/s}$$
- Displacement:
$$d_{3x} = 90.71 \times 1800 = 163278\ \mathrm{m}$$
$$d_{3y} = -70.71 \times 1800 = -127278\ \mathrm{m}$$
5. **Calculate total displacement components:**
$$D_x = d_{1x} + d_{2x} + d_{3x} = 127284 + 540000 + 163278 = 830562\ \mathrm{m}$$
$$D_y = d_{1y} + d_{2y} + d_{3y} = 127284 + 0 - 127278 = 6\ \mathrm{m}$$
6. **Calculate magnitude of total displacement:**
$$D = \sqrt{D_x^2 + D_y^2} = \sqrt{830562^2 + 6^2} \approx 830562\ \mathrm{m} = 830.6\ \mathrm{km}$$
7. **Calculate total time:**
$$T = 1200 + 3000 + 1800 = 6000\ \mathrm{s}$$
8. **Calculate average ground speed:**
$$v_{avg} = \frac{D}{T} = \frac{830562}{6000} = 138.43\ \mathrm{m/s}$$
**Final answers:**
- Total ground displacement magnitude $\approx 831\ \mathrm{km}$ (Choice B)
- Average ground speed $\approx 138.4\ \mathrm{m/s}$ (Choice B)