Motion Graphs Worksheet-4 | Download Free PDF | Velocity ... - Free Printable

Problem Analysis:

The given problem involves analyzing a velocity-time graph of a fly's motion. The graph shows the velocity of the fly as a function of time, with positive velocities indicating motion to the right and negative velocities indicating motion to the left. We will solve each part step by step.

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Graph Analysis:

- Time axis: Horizontal axis (in seconds).
- Velocity axis: Vertical axis (in meters per second, m/s).
- Key observations:
- Negative velocities indicate motion to the left.
- Positive velocities indicate motion to the right.
- Slope of the graph represents acceleration.
- Area under the curve represents displacement.

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Solution to Each Part:

#### a. Identify section(s) where the fly moves with constant velocity.
- Constant velocity is represented by a horizontal line on the velocity-time graph.
- From the graph:
- Between 5 s to 10 s, the velocity is constant at approximately \(10 \, \text{m/s}\) (positive, moving right).
- Between 10 s to 15 s, the velocity is constant at approximately \(15 \, \text{m/s}\) (positive, moving right).

Answer: Sections 5 s to 10 s and 10 s to 15 s.

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#### b. Identify section(s) where the fly moves right slowing down.
- Moving right corresponds to positive velocities.
- Slowing down means the velocity is decreasing (negative slope).
- From the graph:
- Between 15 s to 20 s, the velocity is positive but decreasing from \(15 \, \text{m/s}\) to \(0 \, \text{m/s}\).

Answer: Section 15 s to 20 s.

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#### c. Identify section(s) where the fly moves left speeding up.
- Moving left corresponds to negative velocities.
- Speeding up means the magnitude of velocity is increasing (steeper negative slope).
- From the graph:
- Between 0 s to 2 s, the velocity is negative and becoming more negative (speeding up to the left).

Answer: Section 0 s to 2 s.

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#### d. When is the fly at rest?
- The fly is at rest when the velocity is zero.
- From the graph:
- At 2 s, the velocity crosses zero (changing from negative to positive).
- At 20 s, the velocity crosses zero again (changing from positive to negative).

Answer: At 2 s and 20 s.

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#### e. What is the average velocity of the fly between 0 and 15 seconds?
- Average velocity is given by:
$$
\text{Average velocity} = \frac{\text{Total displacement}}{\text{Total time}}
$$
- Displacement is the area under the velocity-time graph.
- From 0 s to 15 s:
- 0 s to 2 s: Area is a triangle below the x-axis (negative displacement).
$$
\text{Area} = \frac{1}{2} \times 2 \times 10 = 10 \, \text{m} \quad (\text{negative})
$$
- 2 s to 5 s: Area is a rectangle above the x-axis (positive displacement).
$$
\text{Area} = 3 \times 10 = 30 \, \text{m}
$$
- 5 s to 10 s: Area is a rectangle above the x-axis (positive displacement).
$$
\text{Area} = 5 \times 10 = 50 \, \text{m}
$$
- 10 s to 15 s: Area is a rectangle above the x-axis (positive displacement).
$$
\text{Area} = 5 \times 15 = 75 \, \text{m}
$$
- Total displacement:
$$
\text{Total displacement} = -10 + 30 + 50 + 75 = 145 \, \text{m}
$$
- Total time: \(15 \, \text{s}\)
- Average velocity:
$$
\text{Average velocity} = \frac{145}{15} \approx 9.67 \, \text{m/s}
$$

Answer: \( \boxed{9.67 \, \text{m/s}} \).

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#### f. What is the distance traveled by the fly in this time interval?
- Distance is the total path length, calculated as the sum of the absolute values of the areas under the graph.
- From 0 s to 15 s:
- 0 s to 2 s: Area = \(10 \, \text{m}\) (absolute value).
- 2 s to 5 s: Area = \(30 \, \text{m}\).
- 5 s to 10 s: Area = \(50 \, \text{m}\).
- 10 s to 15 s: Area = \(75 \, \text{m}\).
- Total distance:
$$
\text{Total distance} = 10 + 30 + 50 + 75 = 165 \, \text{m}
$$

Answer: \( \boxed{165 \, \text{m}} \).

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#### g. What is the average speed of the fly in the same time interval?
- Average speed is the total distance traveled divided by the total time.
- Total distance: \(165 \, \text{m}\)
- Total time: \(15 \, \text{s}\)
- Average speed:
$$
\text{Average speed} = \frac{165}{15} = 11 \, \text{m/s}
$$

Answer: \( \boxed{11 \, \text{m/s}} \).

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#### h. What is the average acceleration of the fly in this time interval?
- Average acceleration is given by:
$$
\text{Average acceleration} = \frac{\Delta v}{\Delta t}
$$
- Initial velocity at \(0 \, \text{s}\): \(-10 \, \text{m/s}\)
- Final velocity at \(15 \, \text{s}\): \(15 \, \text{m/s}\)
- Time interval: \(15 \, \text{s}\)
- Average acceleration:
$$
\text{Average acceleration} = \frac{15 - (-10)}{15} = \frac{25}{15} \approx 1.67 \, \text{m/s}^2
$$

Answer: \( \boxed{1.67 \, \text{m/s}^2} \).

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#### i. What is the total displacement of the fly from 0 to 22 seconds?
- Displacement is the net area under the velocity-time graph.
- From 0 s to 22 s:
- 0 s to 2 s: Area = \(-10 \, \text{m}\)
- 2 s to 5 s: Area = \(30 \, \text{m}\)
- 5 s to 10 s: Area = \(50 \, \text{m}\)
- 10 s to 15 s: Area = \(75 \, \text{m}\)
- 15 s to 20 s: Area is a triangle above the x-axis (positive displacement).
$$
\text{Area} = \frac{1}{2} \times 5 \times 15 = 37.5 \, \text{m}
$$
- 20 s to 22 s: Area is a triangle below the x-axis (negative displacement).
$$
\text{Area} = \frac{1}{2} \times 2 \times 10 = 10 \, \text{m} \quad (\text{negative})
$$
- Total displacement:
$$
\text{Total displacement} = -10 + 30 + 50 + 75 + 37.5 - 10 = 172.5 \, \text{m}
$$

Answer: \( \boxed{172.5 \, \text{m}} \).

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Final Answers:

1. Constant velocity: 5 s to 10 s and 10 s to 15 s
2. Moving right slowing down: 15 s to 20 s
3. Moving left speeding up: 0 s to 2 s
4. At rest: 2 s and 20 s
5. Average velocity (0 to 15 s): \( \boxed{9.67 \, \text{m/s}} \)
6. Distance traveled (0 to 15 s): \( \boxed{165 \, \text{m}} \)
7. Average speed (0 to 15 s): \( \boxed{11 \, \text{m/s}} \)
8. Average acceleration (0 to 15 s): \( \boxed{1.67 \, \text{m/s}^2} \)
9. Total displacement (0 to 22 s): \( \boxed{172.5 \, \text{m}} \)