Speed-time graph depicting an athlete's motion during a race, used for calculating distance and acceleration.

Problem Analysis:

The given problem involves analyzing a speed-time graph of an athlete during a race. The graph shows the athlete's speed (in meters per second, m/s) as a function of time (in seconds). We need to extract specific information from the graph and answer the questions accordingly.

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Part (i): Using the graph, determine:

#### (a) The MAXIMUM speed
- The maximum speed is the highest point on the graph.
- From the graph, the speed reaches a peak value of 12 m/s at approximately \( t = 6 \) seconds.
- Therefore, the maximum speed is:
\[
\boxed{12 \text{ m/s}}
\]

#### (b) The number of seconds for which the speed was constant
- The speed is constant when the graph is a horizontal line.
- From the graph, the speed remains constant between \( t = 6 \) seconds and \( t = 10 \) seconds.
- The duration for which the speed is constant is:
\[
10 - 6 = 4 \text{ seconds}
\]
- Therefore, the number of seconds for which the speed was constant is:
\[
\boxed{4 \text{ seconds}}
\]

#### (c) The TOTAL distance covered by the athlete during the race
- The total distance covered is the area under the speed-time graph.
- The graph can be divided into three regions for calculating the area:
1. Region 1 (0 to 6 seconds): A triangle with base = 6 seconds and height = 12 m/s.
\[
\text{Area of triangle} = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 6 \times 12 = 36 \text{ meters}
\]
2. Region 2 (6 to 10 seconds): A rectangle with width = 4 seconds and height = 12 m/s.
\[
\text{Area of rectangle} = \text{width} \times \text{height} = 4 \times 12 = 48 \text{ meters}
\]
3. Region 3 (10 to 13 seconds): A triangle with base = 3 seconds and height = 12 m/s.
\[
\text{Area of triangle} = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 3 \times 12 = 18 \text{ meters}
\]
- Adding the areas of all three regions:
\[
\text{Total distance} = 36 + 48 + 18 = 102 \text{ meters}
\]
- Therefore, the total distance covered by the athlete is:
\[
\boxed{102 \text{ meters}}
\]

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Part (ii): During which time-period of the race was:

#### (a) The speed of the athlete increasing
- The speed increases when the graph has a positive slope.
- From the graph, the speed increases from \( t = 0 \) seconds to \( t = 6 \) seconds.
- Therefore, the speed of the athlete was increasing during:
\[
\boxed{0 \text{ to } 6 \text{ seconds}}
\]

#### (b) The speed of the athlete decreasing
- The speed decreases when the graph has a negative slope.
- From the graph, the speed decreases from \( t = 10 \) seconds to \( t = 13 \) seconds.
- Therefore, the speed of the athlete was decreasing during:
\[
\boxed{10 \text{ to } 13 \text{ seconds}}
\]

#### (c) The acceleration of the athlete zero
- Acceleration is zero when the speed is constant (i.e., the graph is a horizontal line).
- From the graph, the speed is constant from \( t = 6 \) seconds to \( t = 10 \) seconds.
- Therefore, the acceleration of the athlete was zero during:
\[
\boxed{6 \text{ to } 10 \text{ seconds}}
\]

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Final Answers:

1. Part (i):
- (a) Maximum speed: \(\boxed{12 \text{ m/s}}\)
- (b) Number of seconds for constant speed: \(\boxed{4 \text{ seconds}}\)
- (c) Total distance covered: \(\boxed{102 \text{ meters}}\)

2. Part (ii):
- (a) Speed increasing: \(\boxed{0 \text{ to } 6 \text{ seconds}}\)
- (b) Speed decreasing: \(\boxed{10 \text{ to } 13 \text{ seconds}}\)
- (c) Acceleration zero: \(\boxed{6 \text{ to } 10 \text{ seconds}}\)