Velocity-time graph illustrating an athlete's speed changes during a race, including acceleration, constant speed, and deceleration phases, with accompanying questions and solutions.

Let’s solve each question step by step using the velocity-time graph.

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Question 1: Calculate the acceleration of the athlete during the first 4 seconds

Acceleration = change in velocity ÷ time taken

From the graph:
- At time = 0 s, velocity = 0 m/s
- At time = 4 s, velocity = 8 m/s

So,

Acceleration = (8 m/s - 0 m/s) ÷ 4 s
= 8 ÷ 4
= 2 m/s²

✔ This matches the solution given — good!

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Question 2: What was the athlete doing between the 4th and 14th second?

Look at the graph from t = 4 s to t = 14 s. The line is flat (horizontal) at 8 m/s.

That means the velocity is not changing — it’s constant.

So, the athlete is moving at a constant speed of 8 m/s.

No acceleration or deceleration — just steady running.

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Question 3: Calculate the deceleration in the final stage of the race

Final stage = from t = 15 s to t = 16 s (since velocity drops from 8 m/s to 0 m/s)

Deceleration = change in velocity ÷ time
But since it’s slowing down, we’ll get a negative value — but often “deceleration” is reported as positive magnitude.

Change in velocity = final velocity - initial velocity
= 0 m/s - 8 m/s = -8 m/s

Time taken = 16 s - 15 s = 1 s

So,

Deceleration = (-8 m/s) ÷ 1 s = -8 m/s²

But if they ask for “deceleration” as a positive number (how much it slowed), then it’s 8 m/s²

In physics, deceleration is often written as negative acceleration, but sometimes questions want the magnitude. Let’s check context.

The problem says “calculate the deceleration”, and in many school contexts, they expect the positive value when saying “deceleration”.

Looking back at Q1, they wrote acceleration as 2 m/s² (positive). So here, since it’s slowing down, we can say:

Deceleration = 8 m/s² (magnitude)

But to be precise, let’s write it as:

Acceleration = -8 m/s² → so deceleration is 8 m/s²

✔ Final answer: 8 m/s²

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Question 4: How far has the athlete moved in the first 4 seconds?

Distance = area under the velocity-time graph

From 0 to 4 seconds, the shape is a triangle.

Area of triangle = ½ × base × height

Base = 4 s
Height = 8 m/s

Distance = ½ × 4 × 8 = 2 × 8 = 16 meters

✔ Answer: 16 m

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Question 5: What is the total distance travelled by the athlete?

We need to find the total area under the entire graph from t = 0 to t = 16 s.

Break it into parts:

1. From 0 to 4 s → triangle → area = 16 m (from Q4)

2. From 4 to 15 s → rectangle
Width = 15 - 4 = 11 s
Height = 8 m/s
Area = 11 × 8 = 88 m

Wait — look again at the graph.

At t = 15 s, velocity starts dropping? Actually, looking at the graph description:

The graph goes up to 4 s (triangle), then flat until 15 s? Wait — the x-axis shows up to 16 s, and the drop happens between 15 and 16? But in the original image, the flat part ends at 15 s? Let me recheck.

Actually, in the user’s image description, the graph shows:

- Rising from (0,0) to (4,8)
- Flat from (4,8) to (15,8)? Or (14,8)? Wait — the discussion question #2 says “between 4th and 14th second”

And solution #2 says “travelling at constant speed of 8 m/s” — implying from 4 to 14 s.

Then from 14 to 16 s, it drops to zero.

Check the x-axis labels: 0, 2, 4, 6, 8, 10, 12, 14, 16

So likely:

- 0–4 s: rising
- 4–14 s: flat (10 seconds)
- 14–16 s: falling (2 seconds)

Yes — because question #2 says “between 4th and 14th second”, and question #3 says “final stage” which must be 14–16 s.

Also, in the solutions provided, only Q1 and Q2 are shown, but we can infer.

So correct breakdown:

Part 1: 0–4 s → triangle → area = ½ × 4 × 8 = 16 m

Part 2: 4–14 s → rectangle → width = 10 s, height = 8 m → area = 10 × 8 = 80 m

Part 3: 14–16 s → triangle (falling) → base = 2 s, height = 8 m → area = ½ × 2 × 8 = 8 m

Total distance = 16 + 80 + 8 = 104 meters

✔ Double-check:

- Triangle 1: 16 m
- Rectangle: 10 s × 8 m/s = 80 m
- Triangle 2: ½ × 2 × 8 = 8 m
- Total: 16+80=96; 96+8=104 m

Perfect.

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Now, compiling all answers:

1. Acceleration first 4 s: 2 m/s²
2. Between 4th and 14th second: moving at constant speed of 8 m/s
3. Deceleration in final stage: 8 m/s² (magnitude)
4. Distance in first 4 s: 16 m
5. Total distance: 104 m

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Final Answer:
1. 2 m/s²
2. Moving at a constant speed of 8 m/s
3. 8 m/s²
4. 16 m
5. 104 m