The instantaneous rate of change of a function is given by its derivative evaluated at the given point. In this case, we need to find the derivative of $f(t) = 2t - 9$ and then evaluate it at $t = 1$.

1. Differentiate $f(t)$ with respect to $t$: $f'(t) = \frac{d}{dt}(2t - 9) = 2$

2. Now, evaluate the derivative at $t = 1$: $f'(1) = 2$

Thus, the instantaneous rate of change of the position function at $t = 1$ is 2 feet per second.

Would you like more details on this process, or do you have any other questions?

Here are 5 related questions to explore:

1. How do you interpret the rate of change in a real-world scenario?
2. What is the significance of the derivative in position-time functions?
3. How would the rate of change differ if the position function had a quadratic term?
4. Can the rate of change ever be negative for certain functions?
5. How do you find the rate of change at a general time $t = a$?

Tip: For linear functions like $f(t) = 2t - 9$, the rate of change (derivative) is constant across all $t$.