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Let's go over the problem step by step.

### Problem Overview:
A cyclist participates in a 100-kilometer race and passes the 70-kilometer mark after 2 hours, traveling at a velocity of 45 km/h.

### Part (A):
**Find the cyclist's average velocity during the first two hours of the race.**

The cyclist traveled 70 kilometers in 2 hours, so the average velocity \(v_{avg}\) can be calculated using the formula:

\[
v_{avg} = \frac{\text{distance}}{\text{time}} = \frac{70 \, \text{km}}{2 \, \text{hours}} = 35 \, \text{km/h}
\]

Thus, the average velocity during the first two hours is **35 km/h**.

### Part (B):
**Let \( f(x) \) represent the distance traveled (in kilometers) as a function of time \(x\) (in hours). Find the slope of the secant line through the points \( (0, f(0)) \) and \( (2, f(2)) \) on the graph of \( y = f(x) \).**

- \( f(0) \) represents the distance traveled at time \(x = 0\), which is **0 km**.
- \( f(2) \) represents the distance traveled at time \(x = 2\), which is **70 km**.

The slope of the secant line is the change in distance over the change in time:

\[
\text{slope} = \frac{f(2) - f(0)}{2 - 0} = \frac{70 - 0}{2} = 35 \, \text{km/h}
\]

Thus, the slope of the secant line is **35 km/h**.

### Part (C):
**Find the equation of the tangent line to the graph of \( y = f(x) \) at the point \( (2, f(2)) \).**

Since the cyclist is traveling at a velocity of 45 km/h at the 2-hour mark, the slope of the tangent line at \( x = 2 \) is **45** km/h.

Using the point-slope form of a line equation:

\[
y - f(2) = m(x - 2)
\]

where \( m = 45 \), \( f(2) = 70 \), and \( x = 2 \), we get:

\[
y - 70 = 45(x - 2)
\]

Simplifying:

\[
y = 45x - 90 + 70
\]
\[
y = 45x - 20
\]

Thus, the equation of the tangent line is:

\[
y = 45x - 20
\]

Let me know if you have any questions or want more details!

### Relative Questions:
1. How would the average velocity change if the cyclist covered 80 kilometers in 2 hours instead?
2. What would be the slope of the secant line if the cyclist traveled 60 kilometers in 2 hours?
3. How does the slope of the secant line relate to the average velocity of the cyclist?
4. If the cyclist maintained a constant velocity, what would be the equation of the distance function \( f(x) \)?
5. How does the tangent line at a point on a distance-time graph represent the cyclist's instantaneous velocity?

### Tip:
When dealing with slopes of secant and tangent lines in motion problems, remember that the secant line gives you the average velocity over a time interval, while the tangent line gives you the instantaneous velocity at a specific point in time.

Two hours after the start of a 100-kilometer bicycle race, a cyclist passes the 70-kilometer mark while riding at a velocity of 45 kilometers per hour. Complete parts (A) through (C) below.

Learn how to calculate the average velocity, secant line slope, and equation of the tangent line for a cyclist covering 70 kilometers in 2 hours during a 100-kilometer race. Understand key concepts such as the distance-time function and instantaneous velocity. Suitable for Grades 9-11.

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