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Let's go through each part of this problem step by step:

### Part (a): Plotting the Graph
- You need to plot the distance \( s \) (in meters) against time \( t \) (in seconds).
- \( t \) is plotted along the x-axis, and \( s \) is plotted along the y-axis.
- From the table, you have the following values:

| \( t \) (s) | 0.0 | 2.5 | 5.5 | 11.0 | 16.0 | 22.0 | 26.5 | 32.0 | 39.5 | 47.5 | 56.0 |
|-------------|-----|-----|-----|------|------|------|------|------|------|------|------|
| \( s \) (m) | 0   | 5   | 10  | 15   | 20   | 25   | 30   | 35   | 40   | 45   | 50   |

- Draw a smooth curve connecting these points.

### Part (b): Calculating Average Speed
Average speed can be calculated using the formula:
\[
\text{Average speed} = \frac{\Delta s}{\Delta t}
\]

#### (i) From 0 to 10 s
- From \( t = 0 \) s to \( t = 10 \) s:
  - \( s(10) = 11.0 \, \text{m} \), \( s(0) = 0 \, \text{m} \)
  - \(\Delta s = 11.0 - 0 = 11.0 \, \text{m}\)
  - \(\Delta t = 10 - 0 = 10 \, \text{s}\)
  - \(\text{Average speed} = \frac{11.0 \, \text{m}}{10 \, \text{s}} = 1.1 \, \text{m/s}\)

#### (ii) From 20 s to 35 s
- From \( t = 20 \) s to \( t = 35 \) s:
  - \( s(35) = 32.0 \, \text{m} \), \( s(20) = 16.0 \, \text{m} \)
  - \(\Delta s = 32.0 - 16.0 = 16.0 \, \text{m}\)
  - \(\Delta t = 35 - 20 = 15 \, \text{s}\)
  - \(\text{Average speed} = \frac{16.0 \, \text{m}}{15 \, \text{s}} = 1.07 \, \text{m/s}\)

#### (iii) The Last 10 s (from 40 s to 50 s)
- From \( t = 40 \) s to \( t = 50 \) s:
  - \( s(50) = 56.0 \, \text{m} \), \( s(40) = 39.5 \, \text{m} \)
  - \(\Delta s = 56.0 - 39.5 = 16.5 \, \text{m}\)
  - \(\Delta t = 50 - 40 = 10 \, \text{s}\)
  - \(\text{Average speed} = \frac{16.5 \, \text{m}}{10 \, \text{s}} = 1.65 \, \text{m/s}\)

#### Suggesting Why Average Speeds Are Different
The average speeds are different because the swimmer's velocity is not constant throughout the race. Early on, the swimmer might be accelerating, whereas later in the race, they might slow down or maintain a steady speed.

### Part (c): Tangent at \( t = 20 \) s
To deduce the swimmer’s instantaneous speed at \( t = 20 \) s:
1. Draw a tangent line to the curve at \( t = 20 \) s.
2. Find the gradient of this tangent line.

The gradient of the tangent will give the instantaneous speed. The steeper the line, the greater the speed at that point.

Would you like me to further explain any of these parts, or help you with graph plotting? Here are some related questions:

1. What is the difference between average speed and instantaneous speed?
2. How does drawing a tangent line help in finding the instantaneous speed?
3. Why does the average speed change at different time intervals?
4. How can acceleration be deduced from a distance-time graph?
5. What is the significance of curvature in a distance-time graph?

**Tip:** When drawing a tangent, ensure it touches the curve at only one point without cutting across it to accurately represent the instantaneous rate of change.

A swimmer dives in and is timed at various points in a 50 m race. The record is shown in the table:
Time intervals are 0, 2.5, 5.5, 11.0, 16.0, 22.0, 26.5, 32.0, 39.5, 47.5, and 56.0 seconds, corresponding to distances of 0, 5, 10, 15, 20, 25, 30, 35, 40, 45, and 50 meters respectively.
(a) Plot a graph of distance vs time.
(b) Calculate the average speed for different time intervals.
(c) Draw a tangent at t = 20 s and find the instantaneous speed.

Learn how to plot a distance-time graph for a swimmer's 50 m race, calculate average speed for various time intervals, and find instantaneous speed by drawing a tangent at t = 20 s. Includes step-by-step solution and important speed-related formulas.

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