To find the average speed for the two-way trip, we need to use the concept of harmonic mean for speeds, as the distance for both trips is the same. The formula for the average speed in such cases is given by:

$\text{Average speed} = \frac{2 \cdot v_1 \cdot v_2}{v_1 + v_2}$

Where:

• $v_1 = 22.6 \, \text{m/s}$ (speed to school)
• $v_2 = 16.8 \, \text{m/s}$ (speed back home)

Step-by-Step Calculation:

$\text{Average speed} = \frac{2 \cdot 22.6 \cdot 16.8}{22.6 + 16.8}$ $\text{Average speed} = \frac{760.32}{39.4}$ $\text{Average speed} \approx 19.29 \, \text{m/s}$

So, the average speed of the student for the two-way trip is approximately 19.29 m/s.

Would you like more details on how this formula is derived, or any clarifications?

Related Questions:

1. What is the time taken for the trip to school?
2. What is the time taken for the trip back home?
3. How does the total distance factor into calculating the average speed?
4. How does the harmonic mean differ from the arithmetic mean in such cases?
5. What would happen to the average speed if the trip back was slower?

Tip: In physics, average speed over equal distances but different speeds requires harmonic mean because time spent on each segment is inversely related to speed.