Let's solve this step-by-step.

Part (a): Finding the Constant Speed for the Second Distance

1. Definition of Average Velocity:
   The average velocity $V_{avg}$ for the entire trip is given by:

   $V_{avg} = \frac{\text{Total Distance}}{\text{Total Time}}$

2. Total Distance and Time for the Trip:
   The car travels two distances of $d$ each (forward in the same direction), so the total distance is:

   $\text{Total Distance} = d + d = 2d$

   Let's denote the speed during the first distance as $v_1 = 44.5 \, \text{mi/h}$, and the speed during the second distance as $v_2$. The time taken to travel each distance is:

   $t_1 = \frac{d}{v_1}, \quad t_2 = \frac{d}{v_2}$

   Therefore, the total time for the trip is:

   $\text{Total Time} = t_1 + t_2 = \frac{d}{v_1} + \frac{d}{v_2}$

3. Substitute Values into the Average Velocity Formula:

   Given that the average velocity for the trip is $29.5 \, \text{mi/h}$, we can write:

   $29.5 = \frac{2d}{\frac{d}{v_1} + \frac{d}{v_2}}$

   Simplify the equation:

   $29.5 = \frac{2d}{\frac{d (v_1 + v_2)}{v_1 v_2}} = \frac{2 v_1 v_2}{v_1 + v_2}$

   Now, substitute $v_1 = 44.5$:

   $29.5 = \frac{2 \cdot 44.5 \cdot v_2}{44.5 + v_2}$

   Let's solve for $v_2$. Cross-multiplying gives:

   $29.5(44.5 + v_2) = 2 \cdot 44.5 \cdot v_2$

   $29.5 \cdot 44.5 + 29.5 v_2 = 89 v_2$

   $29.5 \cdot 44.5 = 89 v_2 - 29.5 v_2$

   $29.5 \cdot 44.5 = 59.5 v_2$

   $v_2 = \frac{29.5 \cdot 44.5}{59.5}$

   Now, compute the value:

   $v_2 \approx 22.05 \, \text{mi/h}$

   So, the constant speed for the second distance $d$ is 22.05 mi/h.

Part (b): Average Velocity for the Trip in Opposite Directions

If the car travels the second distance $d$ in the opposite direction (returning home), the total distance is still $2d$, but the displacement is zero since the car returns to its starting point.

1. Definition of Average Velocity with Opposite Directions:

   Since the car returns to the starting point, the displacement is zero:

   $\text{Displacement} = d - d = 0$

   The average velocity, which depends on displacement, is:

   $V_{avg} = \frac{\text{Displacement}}{\text{Total Time}} = \frac{0}{t_1 + t_2} = 0$

   The magnitude of the average velocity is 0 mi/h.

Summary

• (a) The constant speed for the second distance is 22.05 mi/h.
• (b) The magnitude of the average velocity for the trip in opposite directions is 0 mi/h.

Would you like more details on any part of the solution, or do you have any additional questions?

Related Questions:

1. What is the formula for average speed and how does it differ from average velocity?
2. How do you calculate the time taken to travel a certain distance at a constant speed?
3. What is the significance of displacement in calculating average velocity?
4. How do different speeds affect the average velocity for a round trip?
5. Can the average velocity of a trip be greater than any of the individual speeds?

Tip: Always distinguish between average speed and average velocity; the former considers total distance, while the latter considers displacement.