The position of the particle is given by the equation $x(t) = 10t^2$, where $x$ is in meters and $t$ is in seconds.

To find the average velocity over the time interval from $t = 2.00 \, \text{s}$ to $t = 4.00 \, \text{s}$, we can use the formula for average velocity:

$v_{\text{avg}} = \frac{x(t_2) - x(t_1)}{t_2 - t_1}$ where:

• $t_1 = 2.00 \, \text{s}$
• $t_2 = 4.00 \, \text{s}$

Step 1: Calculate $x(t_1)$ and $x(t_2)$

$x(2) = 10(2)^2 = 10 \times 4 = 40 \, \text{m}$ $x(4) = 10(4)^2 = 10 \times 16 = 160 \, \text{m}$

Step 2: Calculate the average velocity

$v_{\text{avg}} = \frac{x(4) - x(2)}{4 - 2} = \frac{160 - 40}{2} = \frac{120}{2} = 60 \, \text{m/s}$

So, the average velocity of the particle from $t = 2.00 \, \text{s}$ to $t = 4.00 \, \text{s}$ is $60 \, \text{m/s}$.

Would you like a more detailed explanation, or do you have any other questions?

Here are 5 related questions you might want to explore:

1. What is the instantaneous velocity of the particle at $t = 3.00 \, \text{s}$?
2. How would the equation change if the initial position was not zero?
3. What is the total displacement of the particle over the same interval?
4. How does the average velocity compare to the instantaneous velocity at the midpoint of the interval?
5. How does the velocity change if the position function includes a constant term?

Tip: Average velocity is always a measure of the overall change in position over time, while instantaneous velocity measures the rate of change of position at a specific time.