Respuesta :
The displacement and the distance traveled by the particle during the time interval [-2, 5] are -26.83 units and 27.17 units.
Displacement is the straight path between the initial and end positions, whereas distance traveled is the actual length the body covers.
Here, the velocity is given as v(t)=(-t²)+3t-2 and the interval is [-2, 5]. By integrating the given velocity function, we will get the total distance traveled and displacement within the given interval.
The displacement is given as follows,
[tex]\begin{aligned}\int\limits^5_{-2} v(t)dt&=\int\limits^5_{-2} ((-t^2)+3t-2)dt\\&=\left[\frac{-t^3}{3}+\frac{3t^2}{2}-2t\right]^5_{-2}\\&=\left(\frac{5^3}{3}+\frac{3(5)^2}{2}-2(5)\right)-\left(-\frac{(-2)^3}{2}-2(-2)\right)\\&=-\frac{85}{6}-\left(\frac{38}{3}\right)\\&=-\frac{161}{6}\\&=-\mathrm{26.83\;units}\end{aligned}[/tex]
The total distance traveled is given by,
[tex]\begin{aligned}\int\limits^a_b|v(t)|dt&=\int\limits^5_{-2}|(-t^2)+3t-2|dt\\&=\int\limits^1_{-2}(t^2-3t+2)dt+\int\limits^2_1(-t^2-3t+2)dt+\int\limits^5_2(t^2-3t+2)dt\\&=\left[\frac{t^3}{3}-\frac{3t^2}{2}+2t\right]^1_{-2}+\left[-\frac{t^3}{3}+\frac{3t^2}{2}-2t\right]^2_{1}+\left[\frac{t^3}{3}-\frac{3t^2}{2}+2t\right]^5_2\\&=\frac{27}{2}+\frac{1}{6}+\frac{27}{2}\\&=\frac{163}{6}\\&=\mathrm{27.17\;units}\end{aligned}[/tex]
Therefore, the displacement and distance traveled are calculated using the given time interval.
The complete question is -
The velocity function is v(t)=(-t²)+3t-2 for a particle moving along a line. Find the displacement and the distance traveled by the particle during the time interval [-2, 5].
To know more about velocity:
https://brainly.com/question/19979064
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