To find the speed of the falling object between $x = 2$ and $x = 2.5$, you need to compute the average rate of change of the function $f(x) = 5x^2$ over that interval. Then, if needed, you can use the derivative to find the instantaneous velocity at a specific point.

Step 1: Compute the Average Speed

The average speed between $x = 2$ and $x = 2.5$ is given by:

$\frac{f(2.5) - f(2)}{2.5 - 2}$

Calculate function values:

$f(2) = 5(2)^2 = 5(4) = 20$ $f(2.5) = 5(2.5)^2 = 5(6.25) = 31.25$

Compute the average speed:

$\frac{31.25 - 20}{2.5 - 2} = \frac{11.25}{0.5} = 22.5$

Thus, the average speed is 22.5 units per second.

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Step 2: Compute the Instantaneous Speed Using the Derivative

The velocity is the derivative of $f(x)$:

$f'(x) = \frac{d}{dx} (5x^2) = 10x$

To find the instantaneous speed at $x = 2$ and $x = 2.5$:

$f'(2) = 10(2) = 20$ $f'(2.5) = 10(2.5) = 25$

Thus, the instantaneous speeds are:

• 20 units per second at $x = 2$
• 25 units per second at $x = 2.5$

Would you like further clarification on any step?

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Related Questions:

1. How do you find the instantaneous velocity at any given time for this function?
2. How does the derivative represent velocity in real-world motion problems?
3. How would you find the acceleration of the object?
4. What happens to the speed if you extend the interval from $x = 2$ to $x = 3$?
5. How can you interpret the difference between average and instantaneous speed in physics?

Tip:

The instantaneous speed is given directly by the derivative, while the average speed represents the overall change over an interval.