A car travels along a straight line at a constant speed of 44.5 mi/h for a distance d and then another distance d in the same direction at another constant speed. The average velocity for the entire trip is 29.5 mi/h.
(a) What is the constant speed with which the car moved during the second distance d?

(b) Suppose the second distance d were traveled in the opposite direction; you forgot something and had to return home at the same constant speed as found in part (a). What is the average velocity for this trip? (Enter the magnitude.)

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.

Learn how to calculate the constant speed for the second distance and average velocity of a car trip. Solve part (a) and (b) of the problem involving constant speed, average velocity, and time-distance relationships. Suitable for high school students.

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