Constant Velocity Particle Model Worksheet 3: Position vs. Time Graphs

Let’s solve each part of this problem step by step.

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Given Data:

Robin’s position vs. time:

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

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## a. Plot a position vs. time graph for the skater.

To plot this, you would:

- Label the horizontal axis as t (s) — time in seconds.
- Label the vertical axis as x (m) — position in meters.
- Plot each point from the table:
- (0.0, 10.0)
- (1.0, 12.0)
- (2.0, 14.0)
- (5.0, 20.0)
- (8.0, 26.0)
- (10.0, 30.0)

You’ll notice that all these points lie on a straight line. This is because Robin is moving with constant velocity.

✔ Graph Tip: Since the grid is blank, you can choose scale:
- For time (t): Each box = 1 second → 10 boxes cover 0 to 10 s.
- For position (x): Each box = 2 meters → 15 boxes cover 0 to 30 m (but start at 10 m).

Draw a straight line through all the plotted points — it should pass through all of them.

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## b. Explain how you can use the graph to determine how far he was from the origin at t = 6.0 s.

> Note: “From the origin” here likely means from x = 0, not necessarily from his starting point. But since his motion starts at x = 10.0 m, we interpret this as “what is his position x at t = 6.0 s?”

Method using the graph:

1. Locate t = 6.0 s on the horizontal axis.
2. Move vertically up until you hit the straight line you drew.
3. From that point, move horizontally left to read the corresponding x-value on the vertical axis.

This gives you Robin’s position at t = 6.0 s.

✔ Alternative (if graphing is approximate): Since the graph is linear, you can also calculate it mathematically (see part c), but the question asks specifically to use the graph.

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## c. Write a mathematical expression that describes the skater’s motion.

Since the position vs. time graph is a straight line, Robin moves with constant velocity.

The general equation for position under constant velocity is:

> x(t) = x₀ + v·t

Where:
- x(t) = position at time t
- x₀ = initial position (at t=0)
- v = constant velocity
- t = time

From the table:
- At t = 0, x = 10.0 m → so x₀ = 10.0 m

Now find velocity v:

Use any two points to compute slope (velocity):

Example: from t = 0.0 s to t = 1.0 s:
- Δx = 12.0 - 10.0 = 2.0 m
- Δt = 1.0 - 0.0 = 1.0 s
→ v = Δx / Δt = 2.0 m/s

Check another interval: t=2.0 to t=5.0:
- Δx = 20.0 - 14.0 = 6.0 m
- Δt = 3.0 s
→ v = 6.0 / 3.0 = 2.0 m/s ✔

So velocity is constant at 2.0 m/s.

✔ Final Expression:

> x(t) = 10.0 + 2.0·t

*(where x is in meters and t is in seconds)*

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## d. Was his speed constant over the entire interval? How do you know?

✔ Yes, his speed was constant.

How do we know?

1. Graphically: The position vs. time graph is a straight line, which indicates constant velocity (and thus constant speed, since direction isn’t changing — he’s going down the sidewalk in one direction).

2. Mathematically: We calculated the average velocity between multiple intervals, and all gave v = 2.0 m/s. Since velocity doesn’t change, speed (magnitude of velocity) is also constant.

3. Data pattern: Every second, position increases by 2.0 m — consistent rate of change.

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## ✔ Final Answers Summary:

a. Plot the points and draw a straight line through them — all points lie on a straight line.

b. On the graph, find t = 6.0 s on the x-axis, go up to the line, then left to read the x-value (should be 22.0 m).

c. Mathematical expression: x(t) = 10.0 + 2.0·t

d. Yes, speed was constant — shown by straight-line graph and constant slope (velocity) across all time intervals.

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Let me know if you’d like me to sketch the graph or show the calculation for t=6.0 s!