𝑋(𝑡) = 8. 00 𝑡
4 − 3. 00 𝑡
3
+ 5. 00 𝑡
2 − 6. 00 𝑡 + 2. 00

 La magnitud de la velocidad en m/s entre t = 2.00s y t = 3.00s

To find the magnitude of the velocity of the function \( X(t) = 8.00t^4 - 3.00t^3 + 5.00t^2 - 6.00t + 2.00 \) between \( t = 2.00 \) seconds and \( t = 3.00 \) seconds, we first need to find the velocity function. The velocity \( v(t) \) is the derivative of the position function \( X(t) \) with respect to time \( t \).

### Step 1: Find the derivative of \( X(t) \)

Given:

\[
X(t) = 8.00t^4 - 3.00t^3 + 5.00t^2 - 6.00t + 2.00
\]

The derivative \( v(t) = \frac{dX}{dt} \) is:

\[
v(t) = \frac{d}{dt}(8.00t^4) - \frac{d}{dt}(3.00t^3) + \frac{d}{dt}(5.00t^2) - \frac{d}{dt}(6.00t) + \frac{d}{dt}(2.00)
\]

Calculating each term's derivative:

\[
v(t) = 32.00t^3 - 9.00t^2 + 10.00t - 6.00
\]

### Step 2: Calculate the velocity at \( t = 2.00 \) and \( t = 3.00 \)

Substituting \( t = 2.00 \) into \( v(t) \):

\[
v(2) = 32.00(2)^3 - 9.00(2)^2 + 10.00(2) - 6.00
\]

\[
v(2) = 32.00 \times 8 - 9.00 \times 4 + 20.00 - 6.00
\]

\[
v(2) = 256.00 - 36.00 + 20.00 - 6.00
\]

\[
v(2) = 234.00 \, \text{m/s}
\]

Now, substituting \( t = 3.00 \) into \( v(t) \):

\[
v(3) = 32.00(3)^3 - 9.00(3)^2 + 10.00(3) - 6.00
\]

\[
v(3) = 32.00 \times 27 - 9.00 \times 9 + 30.00 - 6.00
\]

\[
v(3) = 864.00 - 81.00 + 30.00 - 6.00
\]

\[
v(3) = 807.00 \, \text{m/s}
\]

### Step 3: Calculate the magnitude of the change in velocity

The magnitude of the velocity change between \( t = 2.00 \) and \( t = 3.00 \) is:

\[
|v(3) - v(2)| = |807.00 - 234.00|
\]

\[
|v(3) - v(2)| = 573.00 \, \text{m/s}
\]

### Conclusion

The magnitude of the change in velocity between \( t = 2.00 \) seconds and \( t = 3.00 \) seconds is **573.00 m/s**.

Would you like further details or have any questions?

### Related Questions

1. How do you find the acceleration function for \( X(t) = 8.00t^4 - 3.00t^3 + 5.00t^2 - 6.00t + 2.00 \)?
2. What is the physical significance of the derivative of a position function?
3. How do you interpret the magnitude of the velocity change in real-world applications?
4. How would you find the average velocity between \( t = 2.00 \) seconds and \( t = 3.00 \) seconds?
5. What would the displacement be between \( t = 2.00 \) seconds and \( t = 3.00 \) seconds?

### Tip

Always remember to take the derivative of each term in the function individually and then combine the results to find the overall derivative.

Calculate the magnitude of velocity change between t = 2.00s and t = 3.00s for the function X(t) = 8.00t^4 - 3.00t^3 + 5.00t^2 - 6.00t + 2.00. Detailed step-by-step solution including velocity function derivation and magnitude calculation.

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