The problem provides the position function of an object moving along a line:

$y = f(x) = 8x^2 - 5x$

where $y$ is the position (in feet), and $x$ is time (in seconds). We are asked to:

1. Find the instantaneous velocity function, $v(x) = f'(x)$.
2. Find the velocity at times $x = 1$, $x = 3$, and $x = 5$.

Step 1: Find the instantaneous velocity function

The instantaneous velocity function is the derivative of the position function with respect to time, $x$.

$f(x) = 8x^2 - 5x$

To find $f'(x)$, we apply the power rule:

$f'(x) = \frac{d}{dx} (8x^2) - \frac{d}{dx} (5x)$ $f'(x) = 16x - 5$

Thus, the instantaneous velocity function is:

$v(x) = 16x - 5$

Step 2: Find the velocity at specific times

Now we will calculate the velocity at $x = 1$, $x = 3$, and $x = 5$.

For $x = 1$: $v(1) = 16(1) - 5 = 16 - 5 = 11 \, \text{feet per second}$

For $x = 3$: $v(3) = 16(3) - 5 = 48 - 5 = 43 \, \text{feet per second}$

For $x = 5$: $v(5) = 16(5) - 5 = 80 - 5 = 75 \, \text{feet per second}$

Final Answers:

• The velocity at $x = 1$ second is $11 \, \text{feet per second}$.
• The velocity at $x = 3$ seconds is $43 \, \text{feet per second}$.
• The velocity at $x = 5$ seconds is $75 \, \text{feet per second}$.

Do you have any questions, or would you like more details on any part of the solution?

Here are five related questions you might find useful:

1. What is the acceleration function for the object, given the velocity function?
2. How would the motion of the object change if the position function had an additional constant term?
3. What does the velocity function tell us about the direction of the object's movement?
4. How can we find the total displacement of the object over a specific time interval?
5. How would you interpret negative values of the velocity function in this context?

Tip: The derivative of a position function is always the velocity function, while the derivative of the velocity function gives the acceleration!