Worksheet titled "Speed-Time Graphs" with four problems, each accompanied by a graph illustrating speed over time for various vehicles and scenarios.

Let's solve each problem step by step using the provided speed-time graphs.

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Problem 1: Train Slowing Down

#### Graph Description:
- The train starts at a speed of 30 m/s and slows down linearly to 0 m/s over 60 seconds.

#### Questions:
1. a) What is the deceleration of the train?
- Deceleration is the rate of change of speed with respect to time. It can be calculated as:
\[
\text{Deceleration} = \frac{\Delta v}{\Delta t} = \frac{v_f - v_i}{t_f - t_i}
\]
Here, \( v_i = 30 \) m/s, \( v_f = 0 \) m/s, \( t_i = 0 \) s, and \( t_f = 60 \) s.
\[
\text{Deceleration} = \frac{0 - 30}{60 - 0} = \frac{-30}{60} = -0.5 \, \text{m/s}^2
\]
The magnitude of deceleration is \( 0.5 \, \text{m/s}^2 \).

- Answer: \( \boxed{0.5 \, \text{m/s}^2} \)

2. b) What is the distance travelled?
- The distance travelled is the area under the speed-time graph. For a straight line from (0, 30) to (60, 0), the area is a triangle.
\[
\text{Distance} = \text{Area of triangle} = \frac{1}{2} \times \text{base} \times \text{height}
\]
Here, the base is 60 seconds and the height is 30 m/s.
\[
\text{Distance} = \frac{1}{2} \times 60 \times 30 = 900 \, \text{m}
\]

- Answer: \( \boxed{900 \, \text{m}} \)

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Problem 2: Car's Journey

#### Graph Description:
- The car accelerates from 0 m/s to 15 m/s in 20 seconds, maintains a constant speed of 15 m/s for 30 seconds, and then decelerates to 0 m/s in 10 seconds.

#### Questions:
1. a) What is the maximum speed of the car?
- The maximum speed is the highest point on the graph, which is 15 m/s.

- Answer: \( \boxed{15 \, \text{m/s}} \)

2. b) What is the initial acceleration?
- Initial acceleration is the slope of the line from 0 to 20 seconds.
\[
\text{Initial acceleration} = \frac{\Delta v}{\Delta t} = \frac{15 - 0}{20 - 0} = \frac{15}{20} = 0.75 \, \text{m/s}^2
\]

- Answer: \( \boxed{0.75 \, \text{m/s}^2} \)

3. c) What is the final deceleration?
- Final deceleration is the slope of the line from 50 to 60 seconds.
\[
\text{Final deceleration} = \frac{\Delta v}{\Delta t} = \frac{0 - 15}{60 - 50} = \frac{-15}{10} = -1.5 \, \text{m/s}^2
\]
The magnitude of deceleration is \( 1.5 \, \text{m/s}^2 \).

- Answer: \( \boxed{1.5 \, \text{m/s}^2} \)

4. d) What is the distance travelled?
- The distance is the area under the graph. This area consists of three parts:
- A triangle from 0 to 20 seconds:
\[
\text{Area}_1 = \frac{1}{2} \times 20 \times 15 = 150 \, \text{m}
\]
- A rectangle from 20 to 50 seconds:
\[
\text{Area}_2 = 30 \times 15 = 450 \, \text{m}
\]
- A triangle from 50 to 60 seconds:
\[
\text{Area}_3 = \frac{1}{2} \times 10 \times 15 = 75 \, \text{m}
\]
- Total distance:
\[
\text{Distance} = 150 + 450 + 75 = 675 \, \text{m}
\]

- Answer: \( \boxed{675 \, \text{m}} \)

5. e) What is the average speed for the whole journey?
- Average speed is the total distance divided by the total time.
\[
\text{Average speed} = \frac{\text{Total distance}}{\text{Total time}} = \frac{675}{60} = 11.25 \, \text{m/s}
\]

- Answer: \( \boxed{11.25 \, \text{m/s}} \)

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Problem 3: Speedboat's Journey

#### Graph Description:
- The speedboat accelerates from 0 m/s to 20 m/s in 2 seconds, maintains a constant speed of 20 m/s from 2 to 6 seconds, and then decelerates to 10 m/s in the next 2 seconds.

#### Questions:
1. a) Find the acceleration between \( t = 0 \) and \( t = 2 \).
- Acceleration is the slope of the line from 0 to 2 seconds.
\[
\text{Acceleration} = \frac{\Delta v}{\Delta t} = \frac{20 - 0}{2 - 0} = \frac{20}{2} = 10 \, \text{m/s}^2
\]

- Answer: \( \boxed{10 \, \text{m/s}^2} \)

2. b) Describe the speedboat's motion between \( t = 2 \) and \( t = 6 \).
- Between \( t = 2 \) and \( t = 6 \), the speedboat is moving at a constant speed of 20 m/s. There is no acceleration or deceleration during this interval.

- Answer: The speedboat is moving at a constant speed of 20 m/s.

3. c) Find the distance travelled in the first 4 seconds.
- The distance is the area under the graph from 0 to 4 seconds.
- A triangle from 0 to 2 seconds:
\[
\text{Area}_1 = \frac{1}{2} \times 2 \times 20 = 20 \, \text{m}
\]
- A rectangle from 2 to 4 seconds:
\[
\text{Area}_2 = 2 \times 20 = 40 \, \text{m}
\]
- Total distance:
\[
\text{Distance} = 20 + 40 = 60 \, \text{m}
\]

- Answer: \( \boxed{60 \, \text{m}} \)

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Problem 4: Car and Motorbike Journey

#### Graph Description:
- Both the car and motorbike start from rest and travel in the same direction. The car accelerates to 30 m/s and maintains that speed, while the motorbike accelerates to 20 m/s and maintains that speed.

#### Questions:
1. a) When are they travelling at the same speed?
- From the graph, the car and motorbike have the same speed at \( t = 10 \) seconds, where both are at 20 m/s.

- Answer: \( \boxed{10 \, \text{s}} \)

2. b) How far apart are they when they are travelling at the same speed?
- At \( t = 10 \) seconds:
- Car's distance: The car accelerates linearly from 0 to 20 m/s in 10 seconds. The distance is the area under the graph from 0 to 10 seconds.
\[
\text{Distance}_{\text{car}} = \frac{1}{2} \times 10 \times 20 = 100 \, \text{m}
\]
- Motorbike's distance: The motorbike also accelerates linearly from 0 to 20 m/s in 10 seconds. The distance is the same as the car's distance.
\[
\text{Distance}_{\text{motorbike}} = \frac{1}{2} \times 10 \times 20 = 100 \, \text{m}
\]
- The difference in distance:
\[
\text{Difference} = \text{Distance}_{\text{car}} - \text{Distance}_{\text{motorbike}} = 100 - 100 = 0 \, \text{m}
\]

- Answer: \( \boxed{0 \, \text{m}} \)

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Final Answers:

1. Train:
- a) \( \boxed{0.5 \, \text{m/s}^2} \)
- b) \( \boxed{900 \, \text{m}} \)

2. Car:
- a) \( \boxed{15 \, \text{m/s}} \)
- b) \( \boxed{0.75 \, \text{m/s}^2} \)
- c) \( \boxed{1.5 \, \text{m/s}^2} \)
- d) \( \boxed{675 \, \text{m}} \)
- e) \( \boxed{11.25 \, \text{m/s}} \)

3. Speedboat:
- a) \( \boxed{10 \, \text{m/s}^2} \)
- b) Constant speed of 20 m/s
- c) \( \boxed{60 \, \text{m}} \)

4. Car and Motorbike:
- a) \( \boxed{10 \, \text{s}} \)
- b) \( \boxed{0 \, \text{m}} \)