Motion Graphs Worksheet | PDF | Velocity | Acceleration - Free Printable

Let's solve each part of the Motion Graphs & Kinematics Worksheet step by step, based on the velocity-time graph provided.

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Given:

- The graph is a velocity vs. time (v-t) graph.
- Positive velocity = moving right.
- Negative velocity = moving left.
- Slope of the graph = acceleration.
- Area under the curve = displacement.
- Distance traveled = total area under the curve (absolute value).
- Average velocity = total displacement / total time.
- Average speed = total distance / total time.
- Average acceleration = change in velocity / change in time.

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We'll analyze the graph section by section:

| Time Interval | Velocity Behavior |
|---------------|-------------------|
| 0–2 s | Decreasing from 0 to -10 m/s → accelerating left |
| 2–4 s | Constant at -10 m/s → constant velocity left |
| 4–6 s | Increasing from -10 to 0 → decelerating left (slowing down), then stops |
| 6–8 s | Increasing from 0 to 10 m/s → accelerating right |
| 8–15 s | Constant at 10 m/s → constant velocity right |
| 15–18 s | Decreasing from 10 to 0 → slowing down while going right |
| 18–20 s | Decreasing from 0 to -5 m/s → accelerating left |
| 20–22 s | Constant at -5 m/s → constant velocity left |

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

Constant velocity → horizontal line on v-t graph.

- 2–4 s: v = -10 m/s (constant left)
- 8–15 s: v = 10 m/s (constant right)
- 20–22 s: v = -5 m/s (constant left)

✔ Answer:
2–4 s, 8–15 s, 20–22 s

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

- Moving right → positive velocity
- Slowing down → velocity decreasing (but still positive)

→ Look for positive velocity with negative slope.

This occurs from 15 to 18 seconds.

✔ Answer:
15–18 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 (becoming more negative)

→ Look for negative velocity with negative slope (since velocity is decreasing further).

- 0–2 s: velocity goes from 0 to -10 m/s → accelerating left (speeding up)
- 18–20 s: velocity goes from 0 to -5 m/s → also speeding up left

But note: 18–20 s starts at 0 and goes to -5 → yes, speeding up left.

So both 0–2 s and 18–20 s are intervals where it's moving left and speeding up.

✔ Answer:
0–2 s and 18–20 s

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d. When is the fly at rest?

At rest → velocity = 0

Look for points where the graph crosses zero or stays at zero.

- At t = 4 s, velocity is 0 (transition point)
- From t = 4 s to t = 6 s, velocity increases from -10 to 0 → not at rest
- Wait: only at t = 4 s and t = 6 s does velocity hit 0?

Wait — let’s check carefully.

From the graph:
- From t = 4 s to t = 6 s, the velocity goes from -10 to 0 → so at t = 6 s, velocity is 0.
- Then from t = 6 s to t = 8 s, it increases from 0 to 10 → so it was at rest only at t = 6 s?

But actually, if it’s changing from negative to positive, it passes through zero at t = 6 s.

But is there any interval where velocity is exactly zero?

No — it only touches zero at t = 6 s and possibly t = 4 s?

Wait — at t = 4 s, velocity is -10 m/s → no.

Let’s recheck:

- t = 0: v = 0
- t = 2 s: v = -10
- t = 4 s: v = -10
- t = 6 s: v = 0
- t = 8 s: v = 10
- ...
- t = 18 s: v = 0
- t = 20 s: v = -5

So velocity is zero at:
- t = 0 s (start)
- t = 6 s
- t = 18 s

But is it at rest during an interval? Only if velocity is zero over a time period.

But here, velocity is only zero at specific points, not over an interval.

So the fly is at rest at t = 0 s, t = 6 s, t = 18 s.

But “when” could mean time(s).

However, sometimes "at rest" means velocity = 0.

So we list the times when v = 0.

But wait — at t = 0, it starts at 0 → but immediately accelerates left → so only instantaneously at rest.

Same at t = 6 and t = 18.

So technically, it is at rest at t = 0 s, t = 6 s, t = 18 s.

But perhaps the question wants intervals.

But since velocity is never zero over an interval, answer is:

✔ Answer:
At t = 0 s, t = 6 s, and t = 18 s

(If the question expects intervals, it would be none — but likely they want the times.)

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e. What is the average velocity of the fly between 0 and 15 seconds?

Average velocity = total displacement / total time

Displacement = area under v-t graph from 0 to 15 s.

We compute area in parts:

#### 1. 0–2 s: triangle below x-axis
- Base = 2 s, height = 10 m/s (downward)
- Area = (1/2)(2)(-10) = -10 m

#### 2. 2–4 s: rectangle below x-axis
- Width = 2 s, height = -10 m/s
- Area = (2)(-10) = -20 m

#### 3. 4–6 s: triangle from -10 to 0
- Base = 2 s, height = 10 m/s (upward from -10 to 0)
- Area = (1/2)(2)(-10 + 0)? Wait — better: it's a triangle from -10 to 0 → area = (1/2)(base)(height) = (1/2)(2)(-10) = -10 m? No!

Wait — from t=4 to t=6, velocity goes from -10 to 0 → so it's a triangle above the negative axis?

Actually, it's a triangle with base 2 s, height 10 m/s, but since it's negative velocity, the area is negative.

But the shape is a triangle rising from -10 to 0 → so area = (1/2)(2)(-10) = -10 m? No.

Wait: area under curve = integral.

The velocity is linear from -10 to 0 → average velocity = (-10 + 0)/2 = -5 m/s

Area = average × time = (-5)(2) = -10 m

Yes.

So:

- 0–2 s: triangle: (1/2)(2)(-10) = -10 m
- 2–4 s: rectangle: (2)(-10) = -20 m
- 4–6 s: triangle: (1/2)(2)(-10) = -10 m? Wait — no!

Wait: from t=4 to t=6, velocity goes from -10 to 0 → so the area is a triangle with base 2, height 10 (magnitude), but since it's above the negative side, it's negative?

Actually, no — the area is negative because velocity is negative.

But the shape is a triangle from -10 to 0 → so the area is:

Area = (1/2) × base × height = (1/2)(2)(-10) = -10 m? That’s incorrect.

Better way: use trapezoid or average.

From t=4 to t=6: v from -10 to 0 → average velocity = (-10 + 0)/2 = -5 m/s

Time = 2 s → displacement = (-5)(2) = -10 m

Yes.

Now:

#### 4. 6–8 s: velocity from 0 to 10 → triangle above x-axis
- Area = (1/2)(2)(10) = +10 m

#### 5. 8–15 s: constant at 10 m/s → rectangle
- Time = 7 s → area = (7)(10) = +70 m

Now sum all displacements from 0 to 15 s:

- 0–2: -10
- 2–4: -20
- 4–6: -10
- 6–8: +10
- 8–15: +70

Total displacement = (-10) + (-20) + (-10) + 10 + 70 = +40 m

Time = 15 s

Average velocity = displacement / time = 40 / 15 = 2.67 m/s

✔ Answer:
2.67 m/s (or 8/3 m/s)

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f. What is the distance traveled by the fly in this time interval?

Distance = total path length = sum of absolute values of areas.

Use same intervals:

- 0–2: | -10 | = 10 m
- 2–4: | -20 | = 20 m
- 4–6: | -10 | = 10 m
- 6–8: | +10 | = 10 m
- 8–15: | +70 | = 70 m

Total distance = 10 + 20 + 10 + 10 + 70 = 120 m

✔ Answer:
120 meters

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g. What is the average speed of the fly in the same time interval?

Average speed = total distance / total time

= 120 m / 15 s = 8.0 m/s

✔ Answer:
8.0 m/s

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h. What is the average acceleration of the fly in this time interval (0 to 15 s)?

Average acceleration = Δv / Δt

Initial velocity at t=0: 0 m/s

Final velocity at t=15: 10 m/s (from graph)

Δv = 10 - 0 = 10 m/s

Δt = 15 s

Average acceleration = 10 / 15 = 0.67 m/s²

✔ Answer:
0.67 m/s²

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i. What is the total displacement of the fly from 0 to 22 seconds?

We already have displacement from 0 to 15 s: +40 m

Now compute from 15 to 22 s:

#### 15–18 s: velocity from 10 to 0 → triangle above x-axis
- Area = (1/2)(3)(10) = +15 m

#### 18–20 s: velocity from 0 to -5 → triangle below x-axis
- Area = (1/2)(2)(-5) = -5 m

#### 20–22 s: constant at -5 m/s → rectangle
- Time = 2 s → area = (2)(-5) = -10 m

Total displacement from 15 to 22 s = 15 - 5 - 10 = 0 m

So total displacement from 0 to 22 s = 40 + 0 = 40 m

✔ Answer:
40 meters

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✔ Final Answers:

a. 2–4 s, 8–15 s, 20–22 s
b. 15–18 s
c. 0–2 s and 18–20 s
d. At t = 0 s, t = 6 s, and t = 18 s
e. 2.67 m/s (or 8/3 m/s)
f. 120 m
g. 8.0 m/s
h. 0.67 m/s²
i. 40 m

Let me know if you'd like a visual breakdown or help plotting the graph!