Physics review worksheet featuring a position-time graph and questions to test understanding of motion, displacement, and velocity.

Since I can't view or access uploaded images directly, I can't see the specific position-time graph you're referring to. However, I can guide you through how to solve typical problems related to position-time and velocity-time graphs, based on the structure of your worksheet.

Let’s assume a common scenario for such a problem (as it's often seen in physics):

---

Assumed Position-Time Graph Description:

Suppose the position-time graph shows the following:

- From 0 to 2 seconds: The object moves from 0 m to 6 m (straight line upward).
- From 2 to 5 seconds: The object stays at 6 m (horizontal line).
- From 5 to 7 seconds: The object moves back from 6 m to 0 m (downward straight line).

This is a typical triangular-shaped graph.

Now let's go step by step.

---

1. Describe the motion of the object from:

#### a) 0–2 seconds:
- The graph is a straight line with positive slope.
- This means the object is moving in the positive direction at constant velocity.
- It starts at 0 m and reaches 6 m in 2 seconds → speed = 6 m / 2 s = 3 m/s.

#### b) 2–5 seconds:
- The graph is flat (horizontal line).
- Position does not change → the object is at rest (velocity = 0 m/s).

#### c) 5–7 seconds:
- The graph is a straight line with negative slope.
- Object is moving back toward the origin at constant velocity.
- From 6 m to 0 m in 2 seconds → velocity = -3 m/s.

#### d) 9–12 seconds:
- Wait — if the graph only goes up to 7 seconds, this might be a typo or error.
- If no data is shown beyond 7 s, we cannot describe motion beyond that time.
- Possibly, the graph ends at 7 s, so no motion occurs from 9–12 s unless extended.

> ⚠️ Please confirm the time range on your graph. If the graph ends at 7 s, then 9–12 s is outside the data.

---

2. What is the total distance traveled?

- Distance is the total path length, regardless of direction.
- 0–2 s: 6 m forward
- 2–5 s: 0 m (stopped)
- 5–7 s: 6 m backward

→ Total distance = 6 + 0 + 6 = 12 meters

---

3. What is the total displacement of the object?

- Displacement = final position – initial position
- Starts at 0 m, ends at 0 m (returns to start)
- So, displacement = 0 m

---

4. What is the average velocity of the object?

- Average velocity = total displacement / total time
- Displacement = 0 m
- Time = 7 seconds (assuming motion ends at 7 s)

→ Average velocity = 0 m / 7 s = 0 m/s

---

5. What is the average speed of the object?

- Average speed = total distance / total time
- Distance = 12 m
- Time = 7 s

→ Average speed = 12 m / 7 s ≈ 1.71 m/s

---

6. Create a velocity-time graph from the position-time graph

We now determine velocity in each interval:

| Time Interval | Velocity (m/s) |
|---------------|----------------|
| 0–2 s | +3 m/s |
| 2–5 s | 0 m/s |
| 5–7 s | -3 m/s |

So, the velocity-time graph will look like:

- A horizontal line at +3 m/s from t = 0 to 2 s
- A horizontal line at 0 m/s from t = 2 to 5 s
- A horizontal line at -3 m/s from t = 5 to 7 s

Plot these on the given grid:

```
Velocity (m/s)
^
| ____
| | |
|______|____|______> Time (s)
0 2 5 7
```

(Use the provided grid to draw three horizontal segments.)

---

✔ Summary Answers:

1. Motion description:
- 0–2 s: Moving forward at constant speed
- 2–5 s: At rest
- 5–7 s: Moving backward at constant speed
- 9–12 s: Not defined (if graph ends at 7 s)

2. Total distance: 12 m
3. Displacement: 0 m
4. Average velocity: 0 m/s
5. Average speed: ~1.71 m/s
6. Velocity-time graph: Three horizontal lines at +3, 0, and -3 m/s over respective intervals.

---

📌 Note: If your actual graph differs (e.g., different shape, time range, or values), please describe it in text or redraw it here, and I’ll adjust the solution accordingly.

Would you like me to help you sketch the velocity-time graph or explain any part further?