Worksheet for analyzing position vs. time graphs in physics.

It looks like you're trying to solve a physics problem involving motion, specifically using position vs. time graphs and kinematic equations. However, the image you've uploaded is not visible to me — I can only see the text description and a "PDF" button, which suggests that the actual content of the worksheet isn't rendered.

But based on the visible text, here's what I can infer about the problem and how to solve it:

---

Problem Summary (Based on Text):

The worksheet is titled:
"Constant Velocity Particle Model Worksheet 3: Position vs. Time Graphs"

You're given data for a student walking at various times and positions:

| t (s) | x (m) |
|-------|-------|
| 0.0 | 0 |
| 1.0 | 12.0 |
| 2.0 | 14.0 |
| 3.0 | 20.0 |
| 5.0 | 26.0 |
| 8.0 | 30.0 |
| 10.0 | 30.0 |

There are four tasks:
1. Plot a position vs. time graph.
2. Use the graph to determine how far the student was from the origin at \( t = 6.0 \, \text{s} \).
3. If the student walks back 10 meters, find the new position at \( t = 6.0 \, \text{s} \), and describe the motion.
4. Write a mathematical expression for the student’s motion.
5. Determine if the speed is constant over the interval.

---

Let’s solve each part step-by-step.

---

Step 1: Plotting the Position vs. Time Graph

We plot time \( t \) on the x-axis and position \( x \) on the y-axis.

Points to plot:
- (0, 0)
- (1, 12)
- (2, 14)
- (3, 20)
- (5, 26)
- (8, 30)
- (10, 30)

Plot these points and connect them with straight lines between known data points.

Note: The motion is not constant velocity throughout because the slope (velocity) changes.

---

Step 2: Find Position at \( t = 6.0 \, \text{s} \)

We don’t have data at \( t = 6.0 \, \text{s} \), but we do have:
- At \( t = 5.0 \, \text{s} \), \( x = 26.0 \, \text{m} \)
- At \( t = 8.0 \, \text{s} \), \( x = 30.0 \, \text{m} \)

So, between \( t = 5 \) and \( t = 8 \), the student moves from 26 m to 30 m in 3 seconds.

Assuming constant velocity in this interval (linear motion):

\[
v = \frac{\Delta x}{\Delta t} = \frac{30 - 26}{8 - 5} = \frac{4}{3} \approx 1.33 \, \text{m/s}
\]

Now, from \( t = 5 \) to \( t = 6 \), time elapsed = 1 second.

So, displacement:
\[
\Delta x = v \cdot \Delta t = \frac{4}{3} \times 1 = \frac{4}{3} \approx 1.33 \, \text{m}
\]

Thus, position at \( t = 6 \, \text{s} \):
\[
x = 26 + 1.33 = 27.33 \, \text{m}
\]

✔ So, at \( t = 6.0 \, \text{s} \), the student is approximately 27.3 m from the origin.

---

Step 3: Student Walks Back 10 Meters at \( t = 6.0 \, \text{s} \)

If the student walks back 10 meters from their current position at \( t = 6 \, \text{s} \), then:

New position:
\[
x = 27.33 - 10 = 17.33 \, \text{m}
\]

But now we need to describe the motion during this change.

Since this happens at \( t = 6.0 \, \text{s} \), we assume an instantaneous change in position or a sudden turn around.

So, the motion would be:
- From \( t = 5 \) to \( t = 6 \): moving forward at ~1.33 m/s
- At \( t = 6 \): suddenly turns and walks back 10 m

But since no time is specified for the return, we interpret it as: At \( t = 6 \, \text{s} \), the student is at 17.33 m, and possibly begins moving backward.

So, the new position at \( t = 6 \, \text{s} \) is 17.3 m.

---

Step 4: Mathematical Expression for Motion

We can model this motion in segments, since the velocity changes.

Let’s break it into intervals:

#### Interval 1: \( t = 0 \) to \( t = 1 \)
From (0,0) to (1,12):
\[
v = \frac{12 - 0}{1 - 0} = 12 \, \text{m/s}
\]
So, \( x(t) = 12t \), for \( 0 \leq t \leq 1 \)

#### Interval 2: \( t = 1 \) to \( t = 2 \)
(1,12) to (2,14):
\[
v = \frac{14 - 12}{2 - 1} = 2 \, \text{m/s}
\]
So, \( x(t) = 12 + 2(t - 1) = 2t + 10 \), for \( 1 \leq t \leq 2 \)

Wait — check: at \( t = 1 \): \( 2(1) + 10 = 12 \), good; at \( t = 2 \): \( 4 + 10 = 14 \), good.

#### Interval 3: \( t = 2 \) to \( t = 3 \)
(2,14) to (3,20):
\[
v = \frac{20 - 14}{1} = 6 \, \text{m/s}
\]
\( x(t) = 14 + 6(t - 2) = 6t + 2 \), for \( 2 \leq t \leq 3 \)

Check: \( t = 2 \): \( 12 + 2 = 14 \), \( t = 3 \): \( 18 + 2 = 20 \), good.

#### Interval 4: \( t = 3 \) to \( t = 5 \)
(3,20) to (5,26):
\[
v = \frac{26 - 20}{2} = 3 \, \text{m/s}
\]
\( x(t) = 20 + 3(t - 3) = 3t + 11 \), for \( 3 \leq t \leq 5 \)

Check: \( t = 3 \): \( 9 + 11 = 20 \), \( t = 5 \): \( 15 + 11 = 26 \), good.

#### Interval 5: \( t = 5 \) to \( t = 8 \)
(5,26) to (8,30):
\[
v = \frac{4}{3} \approx 1.33 \, \text{m/s}
\]
\( x(t) = 26 + \frac{4}{3}(t - 5) \), for \( 5 \leq t \leq 8 \)

#### Interval 6: \( t = 8 \) to \( t = 10 \)
(8,30) to (10,30):
No movement → \( x(t) = 30 \), for \( 8 \leq t \leq 10 \)

---

So, the motion is piecewise linear, not described by a single equation.

Therefore, the best answer is:

> The student's motion is described by a piecewise function:
>
> \[
> x(t) =
> \begin{cases}
> 12t & 0 \leq t \leq 1 \\
> 2t + 10 & 1 < t \leq 2 \\
> 6t + 2 & 2 < t \leq 3 \\
> 3t + 11 & 3 < t \leq 5 \\
> 26 + \frac{4}{3}(t - 5) & 5 < t \leq 8 \\
> 30 & 8 < t \leq 10
> \end{cases}
> \]

---

Step 5: Is Speed Constant Over the Entire Interval?

No.

Speed is the magnitude of velocity. Since the velocity changes in every segment (12, 2, 6, 3, 1.33, 0 m/s), the speed is not constant.

For example:
- From 0–1 s: 12 m/s
- From 1–2 s: 2 m/s
- Then increases again

So, speed is NOT constant over the entire interval.

---

✔ Final Answers:

1. Graph: Plot the points and connect with straight lines (piecewise linear).
2. Position at \( t = 6 \, \text{s} \): Approximately 27.3 m.
3. After walking back 10 m: New position = 17.3 m.
4. Expression: Piecewise function as above.
5. Is speed constant? No, it varies.

---

If you can upload the actual image or clarify any missing parts (like whether the walk-back is instantaneous or takes time), I can refine the answer further!