Answered

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```latex
\begin{tabular}{|c|c|c|c|c|c|}
\hline
\begin{tabular}{l}
Number of \\
washers
\end{tabular} & Trial & & \begin{tabular}{l}
[tex]$t_1$[/tex] (s)
\end{tabular} & Time to & \begin{tabular}{l}
[tex]$t_2$[/tex] (s)
\end{tabular} \\
\hline
\multirow{3}{}{\begin{tabular}{l}
1 washer mass [tex]$=$[/tex] \\
[tex]$4.9 g$[/tex]
\end{tabular}} & Trial \#1 & 2.24 & \multirow{3}{}{\begin{tabular}{l}
Average \\
2.23
\end{tabular}} & 3.16 & \multirow{3}{*}{\begin{tabular}{l}
Average \\
3.13
\end{tabular}} \\
\hline & Trial \#2 & 2.21 & & 3.08 & \\
\hline & Trial \#3 & 2.23 & & 3.15 & \\
\hline
\end{tabular}

What is the average velocity of the car over the first [tex]$0.25 \, m$[/tex]? [tex]$\square \, m / s$[/tex]

What is the average velocity of the car over the second [tex]$0.25 \, m$[/tex]? [tex]$\square \, m / s$[/tex]
```

Sagot :

GinnyAnswer
To solve this problem, we need to determine the average velocity of the car over two separate segments, each 0.25 meters long. Here’s the step-by-step breakdown:

### Step 1: Determine Average Times
We have two sets of trial times for each 0.25-meter segment. Let's first find the average time for each set.

#### For the first 0.25 meters:
The times recorded are:
- Trial \#1: 2.24 seconds
- Trial \#2: 2.21 seconds
- Trial \#3: 2.23 seconds

To find the average time ([tex]\( t_1 Avg \)[/tex]) for the first 0.25 meters:
[tex]\[ t_1 Avg = \frac{(2.24 + 2.21 + 2.23)}{3} = \frac{6.68}{3} \approx 2.2267 \text{ seconds} \][/tex]

#### For the second 0.25 meters:
The times recorded are:
- Trial \#1: 3.16 seconds
- Trial \#2: 3.08 seconds
- Trial \#3: 3.15 seconds

To find the average time ([tex]\( t_2 Avg \)[/tex]) for the second 0.25 meters:
[tex]\[ t_2 Avg = \frac{(3.16 + 3.08 + 3.15)}{3} = \frac{9.39}{3} = 3.13 \text{ seconds} \][/tex]

### Step 2: Calculate Average Velocities
To find the average velocity, we use the formula:
[tex]\[ \text{Velocity} = \frac{\text{Distance}}{\text{Time}} \][/tex]
For both segments, the distance covered is 0.25 meters.

#### Average velocity for the first 0.25 meters:
[tex]\[ \text{Velocity}_{t1} = \frac{0.25 \text{ m}}{2.2267 \text{ s}} \approx 0.1123 \text{ m/s} \][/tex]

#### Average velocity for the second 0.25 meters:
[tex]\[ \text{Velocity}_{t2} = \frac{0.25 \text{ m}}{3.13 \text{ s}} \approx 0.0799 \text{ m/s} \][/tex]

### Summary of Results
- The average velocity of the car over the first 0.25 meters is approximately [tex]\(0.1123 \text{ m/s}\)[/tex].
- The average velocity of the car over the second 0.25 meters is approximately [tex]\(0.0799 \text{ m/s}\)[/tex].
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