crubats4b5p

2023-03-25

The velocity function is for a particle moving along a line. What is the displacement (net distance covered) of the particle during the time interval [-3,6]?

reabatebhyk

The area under a velocity curve equals the distance traveled.
${\int }_{-3}^{6}v\left(t\right)dt$
$={\int }_{-3}^{6}-{t}^{2}+3t-2\text{X}dt$
$=-\frac{1}{3}{t}^{3}+\frac{3}{2}{t}^{2}-2t{\mid }_{\left(-3\right)}^{6}$
$=\left(-\frac{1}{3}\left({6}^{3}\right)+\frac{3}{2}\left({6}^{2}\right)-2\left(6\right)\right)-\left(-\frac{1}{3}{\left(-3\right)}^{3}+\frac{3}{2}{\left(-3\right)}^{2}-2\left(-3\right)\right)$
$=114-10.5$
$=103.5$
Thus, ${\int }_{-3}^{6}v\left(t\right)dt=103.5$

inpuctists8f5

The sum of the partial integrals gives the total distance (scalar quantity representing actual path length).
$x={\int }_{-3}^{1}\left(0-\left(-{t}^{2}+3t-2\right)dt+{\int }_{1}^{2}\left(-{t}^{2}+3t-2\right)dt+{\int }_{2}^{6}\left({t}^{2}-3t+2\right)dt$
The magnitude of total displacement (vector quantity representing straight line drawn from start to end of motion) is given by the following integral.
$|\stackrel{\to }{x}|=-{\int }_{-3}^{1}\left({t}^{2}-3t+2\right)dt+{\int }_{1}^{2}\left(-{t}^{2}+3t-2\right)dt-{\int }_{2}^{6}\left({t}^{2}-3t+2\right)dt$

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