Find the instantaneous rate of change of the position function y = f(t) in feet at the given time t in seconds.
f(t) = 2t - 9, t= 1

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 \( t \).

Find the instantaneous rate of change of the position function y = f(t) in feet at the given time t in seconds. f(t) = 2t - 9, t= 1

Learn how to find the instantaneous rate of change of the linear position function f(t) = 2t - 9 at t = 1. This problem involves basic calculus concepts like derivatives and is suitable for students in Grades 10-12. Discover how the derivative of a linear function gives a constant rate of change across all values of t.

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