Question

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?

Answer 1

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:

$D = \int_{a}^{b} s(t) dt$

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

$s(t) = t^3 - 3t^2 - 2$

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

$D = \int_{a}^{b} s(t) dt$

$D = \int_{0}^{5} (t^3 - 3t^2 - 2) dt$

$D = \frac{t^4}{4} - t^3 - 2t|_{t = 0}^{t = 5}$

$D = \left(\frac{5^4}{4} - 5^3 - 2(5)\right) - \left(\frac{0^4}{4} - 0^3 - 2(0)\right)$

$D = 21.25$

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

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