This worksheet helps students practice analyzing velocity-time graphs and plotting position-time data for moving objects.

Let's solve the problems step by step using the given velocity-time graph and data table.

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Part I: Velocity-Time Graph Analysis

We are given a velocity-time graph where:
- The vertical axis is Velocity (km/h).
- The horizontal axis is Time (hours), ranging from 0 to 1.1 hours.

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#### Question 1: Analyzing the Velocity-Time Graph

##### a. At what time was the car stopped?

A car is stopped when its velocity = 0 km/h.

From the graph:
- The velocity drops to 0 km/h at t = 0.7 h and stays at zero until t = 0.8 h.

✔ So, the car was stopped between 0.7 h and 0.8 h.

> Answer: 0.7 to 0.8 hours

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##### b. At what time did the car have the greatest velocity?

The greatest velocity is the highest point on the graph.

From the graph:
- The maximum velocity is 60 km/h, which occurs from t = 0.2 h to t = 0.4 h.

✔ So, the car had the greatest velocity during 0.2 to 0.4 hours.

> Answer: 0.2 to 0.4 hours

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##### c. What was the greatest velocity?

As seen above, the peak velocity is 60 km/h.

> Answer: 60 km/h

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##### d. At what time(s) was the car accelerating?

Acceleration occurs when the velocity is changing, i.e., the slope of the graph is not zero.

- Accelerating means velocity is increasing → positive slope
- Decelerating means velocity is decreasing → negative slope
- But in physics, acceleration includes both speeding up and slowing down.

So, acceleration occurs whenever the slope ≠ 0.

From the graph:
- From 0 to 0.2 h: velocity increases → accelerating
- From 0.4 to 0.7 h: velocity decreases → decelerating (still accelerating, but negative)
- From 0.8 to 1.0 h: velocity increases again → accelerating

Between 0.2–0.4 h: constant velocity → no acceleration
Between 0.7–0.8 h: velocity = 0 → stopped, no acceleration

✔ So, the car was accelerating during:
- 0.0 to 0.2 h
- 0.4 to 0.7 h
- 0.8 to 1.0 h

> Answer: 0.0–0.2 h, 0.4–0.7 h, and 0.8–1.0 h

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##### e. How fast was the car going at 1.0 h?

At t = 1.0 h, look at the graph:

- The velocity is 40 km/h (flat line from 1.0 to 1.1 h).

> Answer: 40 km/h

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##### f. What is the acceleration at 0.9 hr?

Acceleration is the slope of the velocity-time graph.

At t = 0.9 h, the car is on the segment from 0.8 h to 1.0 h, where velocity increases from 0 km/h to 40 km/h.

Let’s calculate the slope:

- Δv = 40 km/h – 0 km/h = 40 km/h
- Δt = 1.0 h – 0.8 h = 0.2 h

$$
a = \frac{\Delta v}{\Delta t} = \frac{40}{0.2} = 200 \text{ km/h}^2
$$

But we should convert this into more standard units if needed, but since the question uses km/h and hours, we can keep it as:

> Answer: 200 km/h²

(Note: This is a large value because we're using km/h and hours — in m/s² it would be much smaller, but for this context, it's acceptable.)

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✔ Summary of Answers for Part 1:

| Question | Answer |
|--------|--------|
| a | 0.7 to 0.8 hours |
| b | 0.2 to 0.4 hours |
| c | 60 km/h |
| d | 0.0–0.2 h, 0.4–0.7 h, 0.8–1.0 h |
| e | 40 km/h |
| f | 200 km/h² |

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Part II: Position-Time Graph & Velocity Calculation

We are given a table of positions for two cars over time:

| Time (min) | Car A (km) | Car B (km) |
|------------|------------|------------|
| 0 | 0 | 2 |
| 2 | 2 | 3.5 |
| 4 | 4 | 5 |
| 6 | 6 | 6.5 |
| 8 | 8 | 8 |

We need to:
1. Graph the position vs. time for both cars.
2. Compute the velocity of each car.

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#### Step 1: Plotting the Graph

We’ll plot Position (km) vs. Time (min) on the same axes.

- For Car A: (0,0), (2,2), (4,4), (6,6), (8,8)
- For Car B: (0,2), (2,3.5), (4,5), (6,6.5), (8,8)

Both are straight lines, so they move at constant velocity.

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#### Step 2: Compute Velocity

Velocity = $ \frac{\Delta \text{position}}{\Delta \text{time}} $

We'll use total change from start to end.

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##### Car A:

- Δposition = 8 km – 0 km = 8 km
- Δtime = 8 min = $ \frac{8}{60} = \frac{2}{15} $ hours

But since time is in minutes, let’s compute velocity in km/min, or convert to km/h.

Let’s do km/h:

$$
v_A = \frac{8 \text{ km}}{8 \text{ min}} = 1 \text{ km/min}
$$

Convert to km/h:
$ 1 \text{ km/min} = 60 \text{ km/h} $

✔ So, Car A velocity = 60 km/h

Alternatively, check with slope:
- From (0,0) to (8,8): rise/run = 8/8 = 1 km/min = 60 km/h

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##### Car B:

- Δposition = 8 km – 2 km = 6 km
- Δtime = 8 min = $ \frac{8}{60} = \frac{2}{15} $ h

$$
v_B = \frac{6 \text{ km}}{8 \text{ min}} = 0.75 \text{ km/min}
$$

Convert to km/h:
$$
0.75 \times 60 = 45 \text{ km/h}
$$

✔ So, Car B velocity = 45 km/h

Check with points:
- From (0,2) to (8,8): Δx = 6 km, Δt = 8 min → 6/8 = 0.75 km/min = 45 km/h

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✔ Final Answers for Part 2:

- Car A velocity: 60 km/h
- Car B velocity: 45 km/h

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📈 Graph Description (since you can't draw here):

On the provided grid:
- Car A: Starts at (0,0), goes through (2,2), (4,4), ..., (8,8). A straight line with slope 1.
- Car B: Starts at (0,2), goes through (2,3.5), (4,5), (6,6.5), (8,8). Also a straight line, but shallower slope.

Car A has a steeper slope → faster speed.

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✔ Final Summary:

#### Part I:
1. a. 0.7 to 0.8 h
b. 0.2 to 0.4 h
c. 60 km/h
d. 0.0–0.2 h, 0.4–0.7 h, 0.8–1.0 h
e. 40 km/h
f. 200 km/h²

#### Part II:
- Car A velocity: 60 km/h
- Car B velocity: 45 km/h

Let me know if you'd like a sketch of the graph or help plotting it!