Worksheet on speed-time graphs with four problems and accompanying graphs for analyzing motion.

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

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Question 1: Train Slowing Down

Graph Description: A straight line from (0, 30) to (60, 0), meaning the train starts at 30 m/s and slows down uniformly to 0 m/s over 60 seconds.

#### a) What is the deceleration of the train?

Deceleration = rate of change of speed = gradient of the graph.

$$
\text{Gradient} = \frac{\Delta v}{\Delta t} = \frac{0 - 30}{60 - 0} = \frac{-30}{60} = -0.5 \, \text{m/s}^2
$$

Since deceleration is the magnitude of negative acceleration:

> ✔ Deceleration = 0.5 m/s²

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#### b) What is the distance travelled?

Distance = Area under the speed-time graph.

This is a triangle with:
- Base = 60 s
- Height = 30 m/s

$$
\text{Area} = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 60 \times 30 = 900 \, \text{m}
$$

> ✔ Distance travelled = 900 meters

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Question 2: Car’s Journey (60 seconds)

Graph Description:
- 0–20 s: Accelerates from 0 to 15 m/s
- 20–50 s: Constant speed at 15 m/s
- 50–60 s: Decelerates to 0 m/s

#### a) Maximum speed of the car?

From the graph, the highest point is 15 m/s.

> ✔ Maximum speed = 15 m/s

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#### b) Initial acceleration?

Acceleration = gradient of the first segment (0 to 20 s):

$$
a = \frac{15 - 0}{20 - 0} = \frac{15}{20} = 0.75 \, \text{m/s}^2
$$

> ✔ Initial acceleration = 0.75 m/s²

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#### c) Final deceleration?

Deceleration during 50–60 s:

$$
a = \frac{0 - 15}{60 - 50} = \frac{-15}{10} = -1.5 \, \text{m/s}^2
$$

Magnitude:
> ✔ Final deceleration = 1.5 m/s²

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#### d) Distance travelled?

Total distance = Area under the graph.

Break into three parts:

1. Triangle (0–20 s):
$$
\frac{1}{2} \times 20 \times 15 = 150 \, \text{m}
$$

2. Rectangle (20–50 s):
$$
30 \times 15 = 450 \, \text{m}
$$

3. Triangle (50–60 s):
$$
\frac{1}{2} \times 10 \times 15 = 75 \, \text{m}
$$

Total distance:
$$
150 + 450 + 75 = 675 \, \text{m}
$$

> ✔ Distance travelled = 675 meters

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#### e) Average speed for the whole journey?

$$
\text{Average speed} = \frac{\text{Total distance}}{\text{Total time}} = \frac{675}{60} = 11.25 \, \text{m/s}
$$

> ✔ Average speed = 11.25 m/s

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Question 3: Speedboat’s Journey

Graph Description:
- 0–2 s: Speed increases from 0 to 22 m/s
- 2–6 s: Constant speed at 22 m/s
- 6–8 s: Speed decreases to 10 m/s
- After 8 s: Constant at 10 m/s

(We'll use grid lines — assume each small square is 1 unit.)

#### a) Acceleration between t = 0 and t = 2

$$
a = \frac{\Delta v}{\Delta t} = \frac{22 - 0}{2 - 0} = \frac{22}{2} = 11 \, \text{m/s}^2
$$

> ✔ Acceleration = 11 m/s²

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#### b) Describe motion between t = 2 and t = 6

Speed is constant at 22 m/s → no acceleration.

> ✔ The speedboat moves at a constant speed of 22 m/s.

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#### c) Distance travelled in first 4 seconds

Split into two parts:

1. 0–2 s: Triangle
$$
\text{Area} = \frac{1}{2} \times 2 \times 22 = 22 \, \text{m}
$$

2. 2–4 s: Rectangle (constant speed)
$$
\text{Area} = 2 \times 22 = 44 \, \text{m}
$$

Total:
$$
22 + 44 = 66 \, \text{m}
$$

> ✔ Distance in first 4 seconds = 66 meters

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Question 4: Car and Motorbike

Graph Description:
- Motorbike: Starts at 0, accelerates to 20 m/s at t = 10 s, then stays constant.
- Car: Starts at 0, accelerates to 30 m/s at t = 30 s, then constant.

Both travel in same direction.

#### a) When are they travelling at the same speed?

Look for intersection point of the two lines.

- Motorbike: reaches 20 m/s at t = 10 s and stays there.
- Car: starts accelerating from 0; we need when its speed = 20 m/s.

Car's acceleration phase: from (0,0) to (30,30)

So slope = $ \frac{30}{30} = 1 \, \text{m/s}^2 $

So speed at time $ t $: $ v = t $

Set $ v = 20 $ → $ t = 20 $ seconds

At t = 20 s, both have speed = 20 m/s

> ✔ They are travelling at the same speed at t = 20 seconds

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#### b) How far apart are they when they are travelling at the same speed?

We need distances travelled by each at t = 20 s.

Motorbike:
- 0–10 s: triangle → $ \frac{1}{2} \times 10 \times 20 = 100 $ m
- 10–20 s: rectangle → $ 10 \times 20 = 200 $ m
- Total = $ 100 + 200 = 300 $ m

Car:
- From t = 0 to t = 20 s: triangle with base 20, height 20
- Area = $ \frac{1}{2} \times 20 \times 20 = 200 $ m

Difference:
$$
300 - 200 = 100 \, \text{m}
$$

> ✔ They are 100 meters apart at t = 20 s

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✔ Final Answers Summary:

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#### Question 1:
a) Deceleration = 0.5 m/s²
b) Distance = 900 m

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#### Question 2:
a) Maximum speed = 15 m/s
b) Initial acceleration = 0.75 m/s²
c) Final deceleration = 1.5 m/s²
d) Distance = 675 m
e) Average speed = 11.25 m/s

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#### Question 3:
a) Acceleration = 11 m/s²
b) Motion: Constant speed of 22 m/s
c) Distance in first 4 s = 66 m

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#### Question 4:
a) Same speed at t = 20 seconds
b) Distance apart = 100 meters

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