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

Beginner2023-03-26Added 6 answers

The area under a velocity curve equals the distance traveled.

${\int}_{-3}^{6}v\left(t\right)dt$

$={\int}_{-3}^{6}-{t}^{2}+3t-2{\text{X}}dt$

$=-\frac{1}{3}{t}^{3}+\frac{3}{2}{t}^{2}-2t{\mid}_{{(-3)}}^{{6}}$

$=\left({-\frac{1}{3}\left({6}^{3}\right)+\frac{3}{2}\left({6}^{2}\right)-2\left(6\right)}\right)-\left({-\frac{1}{3}{(-3)}^{3}+\frac{3}{2}{(-3)}^{2}-2(-3)}\right)$

$=114-10.5$

$=103.5$

Thus, ${\int}_{-3}^{6}v\left(t\right)dt=103.5$

${\int}_{-3}^{6}v\left(t\right)dt$

$={\int}_{-3}^{6}-{t}^{2}+3t-2{\text{X}}dt$

$=-\frac{1}{3}{t}^{3}+\frac{3}{2}{t}^{2}-2t{\mid}_{{(-3)}}^{{6}}$

$=\left({-\frac{1}{3}\left({6}^{3}\right)+\frac{3}{2}\left({6}^{2}\right)-2\left(6\right)}\right)-\left({-\frac{1}{3}{(-3)}^{3}+\frac{3}{2}{(-3)}^{2}-2(-3)}\right)$

$=114-10.5$

$=103.5$

Thus, ${\int}_{-3}^{6}v\left(t\right)dt=103.5$

inpuctists8f5

Beginner2023-03-27Added 5 answers

The sum of the partial integrals gives the total distance (scalar quantity representing actual path length).

$x={\int}_{-3}^{1}(0-(-{t}^{2}+3t-2)dt+{\int}_{1}^{2}(-{t}^{2}+3t-2)dt+{\int}_{2}^{6}({t}^{2}-3t+2)dt$

The magnitude of total displacement (vector quantity representing straight line drawn from start to end of motion) is given by the following integral.

$\left|\overrightarrow{x}\right|=-{\int}_{-3}^{1}({t}^{2}-3t+2)dt+{\int}_{1}^{2}(-{t}^{2}+3t-2)dt-{\int}_{2}^{6}({t}^{2}-3t+2)dt$

$x={\int}_{-3}^{1}(0-(-{t}^{2}+3t-2)dt+{\int}_{1}^{2}(-{t}^{2}+3t-2)dt+{\int}_{2}^{6}({t}^{2}-3t+2)dt$

The magnitude of total displacement (vector quantity representing straight line drawn from start to end of motion) is given by the following integral.

$\left|\overrightarrow{x}\right|=-{\int}_{-3}^{1}({t}^{2}-3t+2)dt+{\int}_{1}^{2}(-{t}^{2}+3t-2)dt-{\int}_{2}^{6}({t}^{2}-3t+2)dt$

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