Complete Question:
Calculate the average speed of a complete round trip in which the outgoing 230 km is covered at 98 km/h, followed by a 1.0-h lunch break, and the return 230 km is covered at 60 km/h. Express your answer to two significant figures and include the appropriate units.
Solution:
Step 1: Calculate the time taken for each part of the trip.
1. Outgoing trip:
- Distance = 230 km
- Speed = 98 km/h
- Time = Distance / Speed = \( \frac{230}{98} \) ≈ 2.3469 h
2. Lunch break:
- Time = 1.0 h
3. Return trip:
- Distance = 230 km
- Speed = 60 km/h
- Time = Distance / Speed = \( \frac{230}{60} \) ≈ 3.8333 h
Step 2: Calculate the total distance and total time.
- Total distance = Outgoing + Return = 230 km + 230 km = 460 km
- Total time = Outgoing time + Lunch break + Return time = 2.3469 h + 1.0 h + 3.8333 h ≈ 7.1802 h
Step 3: Calculate the average speed.
\[ \text{Average speed} = \frac{\text{Total distance}}{\text{Total time}} = \frac{460}{7.1802} \approx 64.06 \, \text{km/h} \]
Final Answer:
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Part B
Calculate the average velocity of a complete round trip in which the outgoing 230 km is covered at 98 km/h, followed by a 1.0-h lunch break, and the return 230 km is covered at 60 km/h. Express your answer to two significant figures and include the appropriate units.
Step 1: Determine Total Displacement
- Outgoing trip: 230 km (from point A to point B)
- Return trip: 230 km (back from point B to point A)
- Total displacement = Final position - Initial position = 0 km
Step 2: Calculate Total Time
1. Outgoing trip time:
\[ t_1 = \frac{230 \text{ km}}{98 \text{ km/h}} ≈ 2.3469 \text{ h} \]
2. Lunch break time:
\[ t_2 = 1.0 \text{ h} \]
3. Return trip time:
\[ t_3 = \frac{230 \text{ km}}{60 \text{ km/h}} ≈ 3.8333 \text{ h} \]
4. Total time:
\[ t_{\text{total}} = t_1 + t_2 + t_3 ≈ 2.3469 + 1.0 + 3.8333 ≈ 7.1802 \text{ h} \]
Step 3: Compute Average Velocity
\[ \text{Average velocity} = \frac{\text{Total displacement}}{\text{Total time}} = \frac{0 \text{ km}}{7.1802 \text{ h}} = 0 \text{ km/h} \]
Final Answer:
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