We are given the equation for acceleration as:

$a = A - Cv^2$

where:

• $a$ is acceleration in m/s²,
• $v$ is velocity in m/s,
• $C = 0.0012 \, \text{m}^{-1}$,
• the cyclist starts from rest and reaches a speed of 100 km/h at the 260-meter mark.

Step 1: Convert velocity to m/s

The final speed is given in km/h, but we need it in m/s. To convert from km/h to m/s:

$v = 100 \, \text{km/h} \times \frac{1000 \, \text{m/km}}{3600 \, \text{s/h}} = 27.78 \, \text{m/s}$

Step 2: Determine the acceleration at the 260-meter mark

The cyclist's acceleration at the 260-meter mark is zero because they are no longer accelerating (having reached a constant speed). So, at $v = 27.78 \, \text{m/s}$, we know that $a = 0$. This allows us to solve for $A$.

Substitute into the equation:

$0 = A - C v^2$

$A = C v^2$

Step 3: Solve for $A$

Substitute the values of $C = 0.0012 \, \text{m}^{-1}$ and $v = 27.78 \, \text{m/s}$:

$A = 0.0012 \times (27.78)^2$

Now, calculate:

$A = 0.0012 \times 772.5$

$A \approx 0.927$

Final Answer:

The value of $A$ is approximately 0.927 m/s².

---

Would you like any further details or clarifications on this solution? Here are some additional questions for further understanding:

1. How does the equation $a = A - Cv^2$ model the deceleration due to air resistance in the context of cycling?
2. Why do we assume $a = 0$ at the 260-meter mark, and how does this relate to the concept of terminal velocity?
3. How would the value of $C$ affect the overall acceleration of the cyclist at different speeds?
4. What would happen to the acceleration if the cyclist's final speed was lower or higher than 100 km/h?
5. How can we use similar models to estimate acceleration in other scenarios, like running or car racing?

Tip: When working with velocity-dependent acceleration models, remember that the acceleration is often reduced by the drag force at higher speeds, which is quadratic in nature.