# As above, a particle has velocity v(t)= 3t^2-2t-1 measured in m/s. Compute total distance traveled over [0,2] in meters?

Apr 21, 2018

The particle travels $2$ meters.

#### Explanation:

Determining distance or displacement from a velocity equation entails integrating $v \left(t\right)$ and evaluating from time point $a$ to time point $b$, IE, the following definite integral:

$d = {\int}_{a}^{b} v \left(t\right) \mathrm{dt}$

Let us first note that distance traveled, unlike displacement, will not be impacted by the direction of the velocity at various points in time.

That is, we won't have to account for where $v \left(t\right)$ is positive or negative before integrating.

Then, we take

$d = {\int}_{0}^{2} \left(3 {t}^{2} - 2 t - 1\right) \mathrm{dt} = \left({t}^{3} - {t}^{2} - t\right) {|}_{0}^{2}$

$= \left({2}^{3} - {2}^{2} - 2\right) = 8 - 4 - 2 = 2$

The particle travels $2$ meters.
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