Solved Motion Graphs & Kinematics Worksheet: 1. The graph | Chegg.com - Free Printable

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:

---

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.

---

Now, let’s go through each part.

---

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.

---

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., negative acceleration.

So look for a segment where:
- Velocity is positive,
- And the graph is sloping downward (decreasing velocity).

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

> ✔ Answer b: The fly moves right and slows down during 15–17 seconds.

---

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 we need:
- Negative velocity,
- And the velocity is decreasing (going more negative) → slope is negative.

Look at:
- From t = 0 s to t = 3 s: velocity goes from 0 to –10 m/s → negative and increasing in magnitude → speeding up left.

Also:
- From t = 17 s to t = 19 s: velocity goes from 0 to –5 m/s → negative, increasing in magnitude → speeding up left.

Wait — but from 17 to 19 s, it's going from 0 to –5, so yes, it's speeding up in the left direction.

But is that speeding up? Yes — speed (magnitude) increases from 0 to 5 m/s, and direction is left.

So two intervals:
- 0–3 s: velocity from 0 to –10 → speeding up left
- 17–19 s: velocity from 0 to –5 → speeding up left

But note: 17–19 s may be only partially shown.

But if from 17 to 19 s, velocity goes from 0 to –5 m/s (linear), then yes.

So:

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

---

d. When is the fly at rest?

At rest → velocity = 0 m/s

Find where the graph crosses the time axis (V = 0).

Check:
- At t = 8 s: velocity reaches 0 (from negative to positive)
- At t = 17 s: velocity reaches 0 (from positive to negative)

Are these points instantaneous?

Yes — the fly is at rest only at the instant when velocity is zero.

But if the graph touches zero and stays there, then it's at rest for an interval.

But in this case:
- At t = 8 s: velocity goes from negative to positive → passes through zero → instantaneous rest
- At t = 17 s: velocity goes from positive to negative → passes through zero → instantaneous rest

But is there any interval where velocity is exactly zero?

Only if the graph flat lines at zero.

From the description, it seems:
- From t = 8 s to t = 12 s, velocity goes from 0 to 10 m/s → so at t = 8 s, it's zero, then increases.

Similarly, at t = 17 s, it's zero, then becomes negative.

So unless there’s a flat segment at zero, the fly is only at rest at specific instants.

But often in such graphs, if the line touches zero but doesn’t stay, it's just momentarily at rest.

So:
- At t = 8 s → velocity = 0 → at rest
- At t = 17 s → velocity = 0 → at rest

But is it at rest for a duration?

No — unless the graph is flat at zero.

In this case, no horizontal segment at V=0.

So:

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

---

✔ Final Answers:

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

---

> 📌 Note: These answers are based on a typical interpretation of such a velocity-time graph. If your actual graph differs slightly (e.g., different times or values), please adjust accordingly. But based on standard worksheets, this should match.