Answer :
Let's break down the problem step by step.
### Given Data
1. Distance traveled over the first part: \( 0.25 \, \text{m} \).
2. Distance traveled over the second part: \( 0.25 \, \text{m} \).
3. Average time to travel \( 0.25 \, \text{m} \): \( 2.23 \, \text{s} \).
4. Average time to travel \( 0.50 \, \text{m} \): \( 3.13 \, \text{s} \).
### Calculations
#### 1. Average velocity over the first 0.25 m
The average velocity \( v_1 \) can be calculated using the formula:
[tex]\[ v_1 = \frac{\text{distance}}{\text{time}} \][/tex]
Substitute the given values:
[tex]\[ v_1 = \frac{0.25 \, \text{m}}{2.23 \, \text{s}} \][/tex]
Therefore,
[tex]\[ v_1 \approx 0.112108 \, \text{m/s} \][/tex]
#### 2. Average velocity over the second 0.25 m
Firstly, let's calculate the time taken to travel the second 0.25 m.
[tex]\[ t_{\text{second } 0.25 \, \text{m}} = \text{total time for } 0.50 \, \text{m} - \text{time for first } 0.25 \, \text{m} \][/tex]
[tex]\[ t_{\text{second } 0.25 \, \text{m}} = 3.13 \, \text{s} - 2.23 \, \text{s} \][/tex]
[tex]\[ t_{\text{second } 0.25 \, \text{m}} = 0.90 \, \text{s} \][/tex]
Now we can calculate the average velocity \( v_2 \) over the second 0.25 m:
[tex]\[ v_2 = \frac{0.25 \, \text{m}}{0.90 \, \text{s}} \][/tex]
So,
[tex]\[ v_2 \approx 0.277778 \, \text{m/s} \][/tex]
### Summary
- The average velocity of the car over the first 0.25 m: \( 0.112108 \, \text{m/s} \).
- The average velocity of the car over the second 0.25 m: [tex]\( 0.277778 \, \text{m/s} \)[/tex].
### Given Data
1. Distance traveled over the first part: \( 0.25 \, \text{m} \).
2. Distance traveled over the second part: \( 0.25 \, \text{m} \).
3. Average time to travel \( 0.25 \, \text{m} \): \( 2.23 \, \text{s} \).
4. Average time to travel \( 0.50 \, \text{m} \): \( 3.13 \, \text{s} \).
### Calculations
#### 1. Average velocity over the first 0.25 m
The average velocity \( v_1 \) can be calculated using the formula:
[tex]\[ v_1 = \frac{\text{distance}}{\text{time}} \][/tex]
Substitute the given values:
[tex]\[ v_1 = \frac{0.25 \, \text{m}}{2.23 \, \text{s}} \][/tex]
Therefore,
[tex]\[ v_1 \approx 0.112108 \, \text{m/s} \][/tex]
#### 2. Average velocity over the second 0.25 m
Firstly, let's calculate the time taken to travel the second 0.25 m.
[tex]\[ t_{\text{second } 0.25 \, \text{m}} = \text{total time for } 0.50 \, \text{m} - \text{time for first } 0.25 \, \text{m} \][/tex]
[tex]\[ t_{\text{second } 0.25 \, \text{m}} = 3.13 \, \text{s} - 2.23 \, \text{s} \][/tex]
[tex]\[ t_{\text{second } 0.25 \, \text{m}} = 0.90 \, \text{s} \][/tex]
Now we can calculate the average velocity \( v_2 \) over the second 0.25 m:
[tex]\[ v_2 = \frac{0.25 \, \text{m}}{0.90 \, \text{s}} \][/tex]
So,
[tex]\[ v_2 \approx 0.277778 \, \text{m/s} \][/tex]
### Summary
- The average velocity of the car over the first 0.25 m: \( 0.112108 \, \text{m/s} \).
- The average velocity of the car over the second 0.25 m: [tex]\( 0.277778 \, \text{m/s} \)[/tex].
