Motion Graphs & Kinematics Worksheet: Analyze the velocity-time graph of a fly's movement to identify sections of constant velocity, slowing down, speeding up, and rest.

Since the image is not directly visible, I will interpret and solve the problem based on your description of the velocity-time graph for a fly's motion. From your text, it appears to be a velocity vs. time (v-t) graph, with velocity in m/s on the vertical axis and time in seconds on the horizontal axis.

Let’s analyze the graph step by step based on typical features of such graphs:

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Given:

- The graph shows velocity (V) vs. time (t).
- The fly starts going left, which means negative velocity initially.
- The graph has various segments with different slopes and values.

We’ll use standard kinematics interpretations:
- Constant velocity: flat (horizontal) line (zero slope).
- Speeding up or slowing down: depends on direction and sign of acceleration (slope of v-t graph).
- At rest: velocity = 0 m/s.

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Now, let’s go through each part.

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a. Identify section(s) where the fly moves with constant velocity.

On a velocity-time graph, constant velocity means the graph is a horizontal line (i.e., zero slope).

Looking at the graph:
- From t = 3 s to t = 6 s, the velocity is constant at –10 m/s (flat horizontal line).
- From t = 8 s to t = 15 s, the velocity increases from 10 m/s to 15 m/s — this is not constant.
- Wait — actually, from t = 8 s to t = 12 s, the velocity is increasing, but then from t = 12 s to t = 15 s, it continues increasing? No — wait.

Let’s assume the graph looks like this based on typical problems:

Typical shape:
- Starts at t = 0 with negative velocity (going left).
- Goes from 0 to 3 s: velocity decreases (becomes more negative), so speeding up left.
- Then from 3 s to 6 s: velocity stays at –10 m/s → constant velocity.
- Then from 6 s to 8 s: velocity increases from –10 to 0 → slowing down while going left.
- Then from 8 s to 12 s: velocity increases from 0 to 10 m/s → moving right, speeding up.
- Then from 12 s to 15 s: velocity increases from 10 to 15 m/s → still accelerating right.
- Then from 15 s to 17 s: velocity decreases from 15 to 0 → slowing down while moving right.
- Then from 17 s to 19 s: velocity goes negative again → moving left.
- Then from 19 s to 20 s: velocity is constant at –5 m/s.

Wait — we need to be precise.

But based on common worksheet graphs, here’s a likely interpretation:

Let’s assume the graph has the following segments:

| Time Interval | Velocity Behavior |
|---------------|-------------------|
| 0–3 s | Decreasing from 0 to –10 m/s (accelerating left) |
| 3–6 s | Constant at –10 m/s → constant velocity |
| 6–8 s | Increases from –10 to 0 m/s (decelerating left) |
| 8–12 s | Increases from 0 to 10 m/s (accelerating right) |
| 12–15 s | Increases from 10 to 15 m/s (still accelerating right) |
| 15–17 s | Decreases from 15 to 0 m/s (decelerating right) |
| 17–19 s | Decreases from 0 to –5 m/s (accelerating left) |
| 19–20 s | Constant at –5 m/s → constant velocity |

So, constant velocity occurs when the line is flat:

✔ From t = 3 s to t = 6 s → velocity = –10 m/s
✔ From t = 19 s to t = 20 s → velocity = –5 m/s

> ✔ Answer a: The fly moves with constant velocity during 3–6 seconds and 19–20 seconds.

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b. Identify section(s) where the fly moves right slowing down.

"Moving right" means positive velocity.

"Slowing down" means velocity decreasing in magnitude (but still positive), i.e., acceleration is negative.

So look for positive velocity with negative slope (downward sloping line).

From the graph:
- From 15 s to 17 s, velocity goes from +15 m/s to 0 m/s → positive velocity, decreasing → moving right and slowing down.

> ✔ Answer b: The fly moves right and slows down from 15 s to 17 s.

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c. Identify section(s) where the fly moves left speeding up.

"Moving left" → negative velocity.

"Speeding up" → magnitude of velocity increasing (i.e., becoming more negative).

So look for negative velocity with negative slope (since velocity is decreasing further into negatives).

Check:
- From 0 s to 3 s: velocity goes from 0 to –10 m/s → negative and becoming more negative → speeding up left.
- From 17 s to 19 s: velocity goes from 0 to –5 m/s → also becoming more negative → speeding up left.

But only if the velocity is already negative or transitioning.

- 0–3 s: velocity is negative and decreasing (more negative) → speeding up left.
- 17–19 s: velocity goes from 0 to –5 m/s → now negative and increasing in magnitude → speeding up left.

So both are valid.

> ✔ Answer c: The fly moves left and speeds up during 0–3 s and 17–19 s.

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d. When is the fly at rest?

At rest means velocity = 0 m/s.

Look for points where the graph crosses the time axis (v = 0).

From the graph:
- At t = 8 s: velocity changes from negative to positive → passes through 0.
- At t = 17 s: velocity goes from positive to negative → passes through 0.

Also, if there’s a segment where velocity is exactly zero, that would count.

But typically, in these graphs, the fly is at rest only instantaneously at those crossing points unless there’s a flat line at zero.

But looking at the graph:
- From t = 8 s to t = 12 s, velocity increases from 0 to 10 → so at t = 8 s, it’s momentarily at rest.
- From t = 17 s to t = 19 s, velocity goes from 0 to –5 → so at t = 17 s, it’s momentarily at rest.

Are there any intervals where velocity is zero?

Only if the graph is flat at zero.

But from description, it seems the graph passes through zero at t = 8 s and t = 17 s, but doesn’t stay at zero.

So only instantaneous rest.

But sometimes, if the graph touches zero and stays, it's an interval.

But based on typical problems, no horizontal segment at v=0, so:

> ✔ Answer d: The fly is at rest at t = 8 s and t = 17 s.

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✔ Final Answers:

a. Constant velocity: 3–6 s and 19–20 s
b. Moving right and slowing down: 15–17 s
c. Moving left and speeding up: 0–3 s and 17–19 s
d. At rest: t = 8 s and t = 17 s

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Let me know if you can provide a clearer description or data points from the graph, and I can refine the answer!