To solve this problem, let's determine the average speed for a round trip, where the speeds for each leg of the journey differ.

Given:

• Speed to the store = $20 \, \text{km/h}$
• Speed returning from the store = $30 \, \text{km/h}$

Step-by-Step Solution:

1. Define the Distance: Let the one-way distance to the store be $d$.

2. Calculate the Time for Each Part of the Trip:

   • Time to the store: $t_1 = \frac{d}{20}$
   • Time back from the store: $t_2 = \frac{d}{30}$

3. Total Distance: The total distance for the round trip is: $\text{Total Distance} = d + d = 2d$

4. Total Time: The total time for the round trip is: $\text{Total Time} = t_1 + t_2 = \frac{d}{20} + \frac{d}{30}$

   To add these, we find a common denominator: $\text{Total Time} = \frac{3d}{60} + \frac{2d}{60} = \frac{5d}{60} = \frac{d}{12}$

5. Average Speed: Average speed is given by: $\text{Average Speed} = \frac{\text{Total Distance}}{\text{Total Time}} = \frac{2d}{\frac{d}{12}} = 2d \times \frac{12}{d} = 24 \, \text{km/h}$

Answer:

The average speed for the round trip is $\mathbf{24 \, \text{km/h}}$.

Would you like further details on the calculation steps?

Related Questions:

1. What is the concept of harmonic mean, and how is it used in average speed problems?
2. How would the average speed change if the return speed was 40 km/h instead of 30 km/h?
3. Why is the average speed not simply the average of 20 km/h and 30 km/h?
4. How can we use the formula for average speed in physics and engineering problems?
5. What happens to the average speed if the distance each way is doubled?

Tip:

For problems involving average speeds over equal distances with different speeds, consider using the harmonic mean rather than the arithmetic mean.