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The problem asks for the average speed of a car traveling between two cities, A and B, with different speeds during the onward and return journeys.

### Key Information:
- Speed from A to B: \( 18 \, \text{km/h} \)
- Speed from B to A: \( 36 \, \text{km/h} \)

### Formula for Average Speed:

The average speed for a round trip when the distances are the same but the speeds differ is given by the harmonic mean of the two speeds:
\[
\text{Average Speed} = \frac{2 \cdot v_1 \cdot v_2}{v_1 + v_2}
\]
where:
- \(v_1 = 18 \, \text{km/h}\) (speed from A to B)
- \(v_2 = 36 \, \text{km/h}\) (speed from B to A)

### Solution:

\[
\text{Average Speed} = \frac{2 \cdot 18 \cdot 36}{18 + 36}
\]

Let me calculate this.The average speed for the entire journey is \( 24 \, \text{km/h} \).

Would you like further details or have any questions?

Here are five related questions you might find helpful:
1. What is the difference between average speed and average velocity?
2. How would the average speed change if the speeds were equal for both trips?
3. How does distance impact the calculation of average speed in different cases?
4. Can you explain why the harmonic mean is used for calculating the average speed in such scenarios?
5. How does the average speed formula change if the return distance is different from the onward journey?

**Tip:** The harmonic mean is used when dealing with rates (like speed) over the same distance.

A car traveled from city A to city B with a speed of 18 km/h and returned to city A with a speed of 36 km/h. What is its average speed for the whole journey?

Learn how to calculate the average speed for a round trip when a car travels from city A to city B at 18 km/h and returns at 36 km/h. This problem involves the use of the harmonic mean formula, key for understanding speed, time, and distance calculations.

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