Physics, 19.09.2019 16:30 Jenniferlam12
Aparticle moves along a line so that its velocity at time t is v(t) = t2 − t − 20 (measured in meters per second). (a) find the displacement of the particle during 1 ≤ t ≤ 10. (b) find the distance traveled during this time period.
Answers: 1
![\int\limits^{-10}_1 {[t^2-t-20]} \, dx](/tpl/images/0243/4234/a0aa9.png)
![[\frac{t^3}{3} -\frac{t^2}{2} -20t]_1^{10}](/tpl/images/0243/4234/550b1.png)
![\frac{1}{3}[10^3-1^3]-\frac{1}{2} [10^2-1^2] -20[10-1]](/tpl/images/0243/4234/0f060.png)
![\frac{1}{3}[1000-1]-\frac{1}{2} [99]-20[9]](/tpl/images/0243/4234/5a773.png)
On interval [5, 10], 
![\int\limits^{10}_1 {v(t)} \, dt = \int\limits^5_1 {[-v(t)} \, dt]+ \int\limits^{10}_5 v(t) \, dt](/tpl/images/0243/4234/9850d.png)
![= \int\limits^5_1 {[-t^2+t+20]} \, dt + \int\limits^{10}_5 {[t^2+t+20]} \, dt](/tpl/images/0243/4234/f5d2a.png)
![=[\frac{-t^3}{3} ]_1^5+[\frac{t^2}{2} ]_1^5+20[t]_1^5+[\frac{t^3}{3}-\frac{t^2}{2} -20t ]_5^{10}](/tpl/images/0243/4234/cf421.png)
. Distance traveled will be 
, while displacement will be 
(note the absolute value is needed because part of the movement MIGHT be backwards, which will add to the distance traveled, ie the number of steps, but does not contribute to displacement - think of a guy running in cirles).
and 
- the particle is moving backwards. The integral becomes:
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Aparticle moves along a line so that its velocity at time t is v(t) = t2 − t − 20 (measured in meter...
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