To solve this problem, we'll need to calculate both the average velocity and the instantaneous velocity.

1. Average Velocity Between $t = 2$ and $t = 3$

The average velocity is calculated using the formula:

$\text{Average Velocity} = \frac{s(t_2) - s(t_1)}{t_2 - t_1}$

where $s(t)$ is the position function, $t_1 = 2$, and $t_2 = 3$.

First, calculate $s(2)$ and $s(3)$:

$s(2) = 14(2) + 17 = 28 + 17 = 45 \text{ km}$

$s(3) = 14(3) + 17 = 42 + 17 = 59 \text{ km}$

Now, plug these values into the average velocity formula:

$\text{Average Velocity} = \frac{59 - 45}{3 - 2} = \frac{14}{1} = 14 \text{ km/h}$

So, the average velocity between $t = 2$ and $t = 3$ is 14 km/h.

2. Instantaneous Velocity at $t = 2.5$

The instantaneous velocity is given by the derivative of the position function $s(t)$ with respect to time $t$:

$v(t) = \frac{ds(t)}{dt}$

Given $s(t) = 14t + 17$, the derivative is:

$v(t) = \frac{d}{dt}(14t + 17) = 14$

Since the derivative is constant, the instantaneous velocity at any time $t$ is 14 km/h. Thus, at $t = 2.5$, the instantaneous velocity is also 14 km/h.

Summary:

• Average Velocity between $t = 2$ and $t = 3$: 14 km/h
• Instantaneous Velocity at $t = 2.5$: 14 km/h

Would you like any further details or explanations?

Here are some related questions:

1. How do you calculate average velocity for a non-linear position function?
2. What is the physical significance of instantaneous velocity?
3. How would the solution change if the position function was quadratic?
4. Can average velocity ever be zero? If so, under what conditions?
5. How does the concept of acceleration relate to velocity?
6. What would be the effect of changing the slope in the linear function on velocity?
7. How can we interpret the constant velocity in real-world terms for this scenario?
8. What methods can be used to find instantaneous velocity if the position function is not differentiable?

Tip: When working with linear position functions, the average and instantaneous velocities will always be the same.