The velocity function is v(t)= –t^2+3t – 2 for a particle moving along a line. What is the displacement (net distance covered) of the particle during the time interval [-3,6]?

Sep 22, 2015

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

Explanation:

The area under a velocity curve is equivalent to the distance covered.

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

$= {\int}_{- 3}^{6} - {t}^{2} + 3 t - 2 \textcolor{w h i t e}{\text{X}} \mathrm{dt}$

$= - \frac{1}{3} {t}^{3} + \frac{3}{2} {t}^{2} - 2 t {|}_{\textcolor{b l u e}{\left(- 3\right)}}^{\textcolor{red}{6}}$

$= \left(\textcolor{red}{- \frac{1}{3} \left({6}^{3}\right) + \frac{3}{2} \left({6}^{2}\right) - 2 \left(6\right)}\right) - \left(\textcolor{b l u e}{- \frac{1}{3} {\left(- 3\right)}^{3} + \frac{3}{2} {\left(- 3\right)}^{2} - 2 \left(- 3\right)}\right)$

$= 114 - 10.5$

$= 103.5$

Sep 22, 2015

The original question is a bit confusing as it implies that displacement and distance is the same thing, which it is not.
I have set up the necessary integration for each different case hereunder.

Explanation:

Total distance (scalar quantity representing actual path length) is given by the sum of the partial integrals
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

Total displacement (vector quantity representing straight line drawn from start to end of motion) is given in magnitude by the following integral
$| \vec{x} | = - {\int}_{- 3}^{1} \left({t}^{2} - 3 t + 2\right) \mathrm{dt} + {\int}_{1}^{2} \left(- {t}^{2} + 3 t - 2\right) \mathrm{dt} - {\int}_{2}^{6} \left({t}^{2} - 3 t + 2\right) \mathrm{dt}$

The graph of the velocity function with time makes it clear why these integrals need to be set up for the vector rules to be obeyed and the definitions to be satisfied.
graph{-x^2+3x-2 [-34.76, 38.3, -21.53, 14.98]}