Worksheet for graphing practice, including data sets and sketches of original and test plots.

Let's solve this Scientific Methods Worksheet 1: Graphing Practice step by step. We'll analyze both data sets, graph the original data, determine how to transform the data so that it forms a straight line, and then derive the mathematical relationship.

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Data Set 1: Volume vs. Pressure

#### Given Data:
| Volume (m³) | Pressure (Pascals) |
|-------------|--------------------|
| 0.1 | 40.0 |
| 0.5 | 8.0 |
| 1.0 | 4.0 |
| 4.0 | 1.0 |
| 5.0 | 0.80 |
| 8.0 | 0.50 |
| 10.0 | 0.40 |

#### Step 1: Plot Original Data (Volume vs. Pressure)

When we plot Pressure (P) vs. Volume (V), we observe that as volume increases, pressure decreases — suggesting an inverse relationship.

Looking at the values:

- At V = 0.1 m³ → P = 40.0 Pa
- At V = 0.5 m³ → P = 8.0 Pa
- At V = 1.0 m³ → P = 4.0 Pa
- At V = 4.0 m³ → P = 1.0 Pa

Check if P × V is constant:
- 0.1 × 40.0 = 4.0
- 0.5 × 8.0 = 4.0
- 1.0 × 4.0 = 4.0
- 4.0 × 1.0 = 4.0
- 5.0 × 0.80 = 4.0
- 8.0 × 0.50 = 4.0
- 10.0 × 0.40 = 4.0

✔ All products are 4.0 Pa·m³, so:

> P × V = constant ⇒ P = k / V

This is Boyle’s Law: pressure inversely proportional to volume.

So, to make a straight-line graph, we should plot Pressure (P) vs. 1/Volume (1/V).

#### Step 2: Modify Data – Use 1/V as Independent Variable

| Volume (m³) | 1/Volume (1/m³) | Pressure (Pa) |
|-------------|------------------|---------------|
| 0.1 | 10.0 | 40.0 |
| 0.5 | 2.0 | 8.0 |
| 1.0 | 1.0 | 4.0 |
| 4.0 | 0.25 | 1.0 |
| 5.0 | 0.20 | 0.80 |
| 8.0 | 0.125 | 0.50 |
| 10.0 | 0.10 | 0.40 |

Now plot P vs. 1/V — this should be a straight line.

#### Step 3: Find Slope and Intercept

We expect:
> P = k × (1/V) → linear form: y = mx + b
where:
- y = P
- x = 1/V
- m = slope = k
- b = 0 (should pass through origin)

Let’s calculate slope using two points:

Take (1/V = 1.0, P = 4.0) and (1/V = 10.0, P = 40.0):

Slope = (40.0 - 4.0) / (10.0 - 1.0) = 36.0 / 9.0 = 4.0 Pa·m³

All other points confirm this:
- At 1/V = 0.25, P = 1.0 → 0.25 × 4 = 1.0 ✔
- At 1/V = 0.10, P = 0.40 → 0.10 × 4 = 0.40 ✔

So the equation is:

> P = 4.0 × (1/V) or P = 4.0 / V

✔ Units:
- Pressure in Pascals (Pa)
- Volume in m³
- Constant: 4.0 Pa·m³

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Data Set 2: Time vs. Position

#### Given Data:
| Time (s) | Position (m) |
|----------|--------------|
| 0.10 | 0.03 |
| 0.20 | 0.12 |
| 1.0 | 3.0 |
| 2.0 | 12.0 |
| 3.0 | 27.0 |
| 4.0 | 48.0 |
| 5.0 | 75.0 |

#### Step 1: Plot Original Data (Time vs. Position)

Plotting position (y-axis) vs. time (x-axis), the curve appears quadratic — increasing faster than linear.

Let’s check if position is proportional to t²:

Compute t² and compare with position:

| t (s) | t² (s²) | Position (m) |
|-------|---------|--------------|
| 0.10 | 0.01 | 0.03 |
| 0.20 | 0.04 | 0.12 |
| 1.0 | 1.0 | 3.0 |
| 2.0 | 4.0 | 12.0 |
| 3.0 | 9.0 | 27.0 |
| 4.0 | 16.0 | 48.0 |
| 5.0 | 25.0 | 75.0 |

Now compute ratio: Position / t²

- 0.03 / 0.01 = 3.0
- 0.12 / 0.04 = 3.0
- 3.0 / 1.0 = 3.0
- 12.0 / 4.0 = 3.0
- 27.0 / 9.0 = 3.0
- 48.0 / 16.0 = 3.0
- 75.0 / 25.0 = 3.0

✔ All ratios equal 3.0 m/s²

So:
> Position = 3.0 × t²

Thus, to get a straight line, plot Position (m) vs. t² (s²)

#### Step 2: Modified Data – Plot Position vs. t²

| t (s) | t² (s²) | Position (m) |
|-------|---------|--------------|
| 0.10 | 0.01 | 0.03 |
| 0.20 | 0.04 | 0.12 |
| 1.0 | 1.0 | 3.0 |
| 2.0 | 4.0 | 12.0 |
| 3.0 | 9.0 | 27.0 |
| 4.0 | 16.0 | 48.0 |
| 5.0 | 25.0 | 75.0 |

Plotting Position vs. t² gives a straight line through the origin.

#### Step 3: Equation of Line

From above:
> Position = 3.0 × t²

So:
- Slope = 3.0 m/s²
- y-intercept = 0

This matches the motion of an object under constant acceleration starting from rest:
> x = (1/2) a t²

Compare:
> x = 3.0 t² ⇒ (1/2)a = 3.0 ⇒ a = 6.0 m/s²

But we don’t need to find acceleration unless asked. The mathematical expression is:

> Position = 3.0 × t²

Units:
- Position: meters (m)
- Time: seconds (s)
- Constant: 3.0 m/s²

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✔ Final Answers

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#### Data Set 1: Volume vs. Pressure

- Original graph: Curved (inverse relationship)
- Modified variable: Plot Pressure vs. 1/Volume
- Straight-line graph: Yes
- Mathematical Expression:
> P = 4.0 / V
where:
- P = pressure in Pascals (Pa)
- V = volume in m³
- 4.0 has units Pa·m³

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#### Data Set 2: Time vs. Position

- Original graph: Parabolic (quadratic)
- Modified variable: Plot Position vs. t²
- Straight-line graph: Yes
- Mathematical Expression:
> x = 3.0 × t²
where:
- x = position in meters (m)
- t = time in seconds (s)
- 3.0 has units m/s²

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📌 Summary of Steps for Each Dataset:

| Data Set | Original Graph Shape | Transformation | Linear Relationship | Equation |
|--------|----------------------|----------------|---------------------|---------|
| 1 | Inverse curve | Plot P vs. 1/V | Straight line | P = 4.0 / V |
| 2 | Quadratic curve | Plot x vs. t² | Straight line | x = 3.0 t² |

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🔍 How to Sketch the Graphs (Instructions)

#### For Data Set 1:
- Original sketch: Plot V on x-axis, P on y-axis → downward-curving hyperbola.
- Test plot: Plot 1/V on x-axis, P on y-axis → straight line through origin with slope 4.0.

#### For Data Set 2:
- Original sketch: Plot t on x-axis, x on y-axis → parabola opening upward.
- Test plot: Plot t² on x-axis, x on y-axis → straight line through origin with slope 3.0.

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✔ Final Answer Boxed:

> Data Set 1:
> $ \boxed{P = \frac{4.0\ \text{Pa}\cdot\text{m}^3}{V}} $

> Data Set 2:
> $ \boxed{x = 3.0\ \frac{\text{m}}{\text{s}^2} \cdot t^2} $

These equations represent the physical relationships observed in the data.