Let's solve this step by step.

(i) Calculate the acceleration of the cyclist during the first 8 seconds.

The acceleration can be calculated using the formula: $a = \frac{\Delta v}{\Delta t}$ From the graph:

• The final speed at $t = 8 \, \text{s}$ is $v = 6 \, \text{m/s}$,
• The initial speed at $t = 0 \, \text{s}$ is $0 \, \text{m/s}$,
• The time interval is $\Delta t = 8 \, \text{s}$.

Substitute the values: $a = \frac{6 - 0}{8} = \frac{6}{8} = 0.75 \, \text{m/s}^2$

(ii) Describe the motion of the cyclist between $t = 8 \, \text{s}$ and $t = 90 \, \text{s}$.

From the graph:

• Between $t = 8 \, \text{s}$ and $t = 90 \, \text{s}$, the speed remains constant at $6 \, \text{m/s}$.
• This indicates uniform motion (constant velocity).

Conclusion: The cyclist is moving at a constant speed of $6 \, \text{m/s}$ during this time interval.

(iii) Find the value of $T$ given the total distance is $558 \, \text{m}$.

The total distance can be calculated as the area under the speed-time graph. The graph consists of three regions:

1. A triangle (acceleration phase) from $t = 0$ to $t = 8 \, \text{s}$,
2. A rectangle (constant speed) from $t = 8 \, \text{s}$ to $t = 90 \, \text{s}$,
3. A triangle (deceleration phase) from $t = 90 \, \text{s}$ to $t = T \, \text{s}$.

Step 1: Area of the first triangle (acceleration phase):

$\text{Area} = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 8 \times 6 = 24 \, \text{m}$

Step 2: Area of the rectangle (constant speed phase):

$\text{Area} = \text{base} \times \text{height} = (90 - 8) \times 6 = 82 \times 6 = 492 \, \text{m}$

Step 3: Area of the second triangle (deceleration phase):

Let $T - 90$ be the base of the triangle, and $6 \, \text{m/s}$ the height: $\text{Area} = \frac{1}{2} \times (T - 90) \times 6$

Step 4: Total distance:

The sum of all areas equals $558 \, \text{m}$: $24 + 492 + \frac{1}{2} \times (T - 90) \times 6 = 558$

Simplify: $24 + 492 + 3(T - 90) = 558$ Combine terms: $516 + 3(T - 90) = 558$ $3(T - 90) = 42$ $T - 90 = 14$ $T = 104 \, \text{s}$

(iv) Convert $6 \, \text{m/s}$ into $\text{km/h}$.

To convert meters per second to kilometers per hour, use the conversion factor: $1 \, \text{m/s} = 3.6 \, \text{km/h}$

Thus: $6 \, \text{m/s} = 6 \times 3.6 = 21.6 \, \text{km/h}$

Final Answers:

1. Acceleration during the first 8 seconds: $0.75 \, \text{m/s}^2$
2. Motion between $t = 8 \, \text{s}$ and $t = 90 \, \text{s}$: Constant speed of $6 \, \text{m/s}$
3. Value of $T$: $104 \, \text{s}$
4. $6 \, \text{m/s}$ converted to km/h: $21.6 \, \text{km/h}$

Do you want me to explain any part in more detail?

Here are 5 related questions:

1. How is the area under a speed-time graph related to distance traveled?
2. How would you calculate the deceleration in the last phase of motion?
3. What is the significance of a horizontal line on a speed-time graph?
4. How do you calculate average speed for the entire journey?
5. How can you determine the total time for a journey if given the graph and distance?

Tip: For speed-time graphs, the total area under the graph always gives the distance traveled.