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The vertical position of a particle is given by the function s = t^3 - 3t^2 - 2. How far does the particle travel between t = 0 and t = 5?

Sagot :

Using integrals, it is found that the particle travels 21.25 units between t = 0 and t = 5.

How is the total distance traveled calculated?

  • The total distance traveled by a particle modeled by an equation of position s(t) between t = a and t = b is given by:

[tex]D = \int_{a}^{b} s(t) dt[/tex]

In this problem, the equation that models the position of the particle is:

[tex]s(t) = t^3 - 3t^2 - 2[/tex]

Hence, applying integral properties, then the Fundamental Theorem of Calculus, we have that the distance traveled is of:

[tex]D = \int_{a}^{b} s(t) dt[/tex]

[tex]D = \int_{0}^{5} (t^3 - 3t^2 - 2) dt[/tex]

[tex]D = \frac{t^4}{4} - t^3 - 2t|_{t = 0}^{t = 5}[/tex]

[tex]D = \left(\frac{5^4}{4} - 5^3 - 2(5)\right) - \left(\frac{0^4}{4} - 0^3 - 2(0)\right)[/tex]

[tex]D = 21.25[/tex]

The particle travels 21.25 units between t = 0 and t = 5.

You can learn more about integrals at https://brainly.com/question/20733870

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