Question

Have I worked this correctly? A particle moves along a line with velocity function v(t)=t^2-t, where v is measured in meters per second. Determine both the displacement and the distance travelled by the particle during the time interval [0,5].

I'll put my work in the comments....

Answer 1

For the Displacement I got this:
\[\int\limits_{0}^{5}(t ^{2}-t)\] = \[\frac{ t^3 }{ 3 }-\frac{ t^2 }{ 2 } \] I then put 5 in for t and evaluated and subtracted 0 because t at 0=0 and came up with displacement = 175/6m

Answer 2

displacement:$$x=\int_0^5v\ dt=\int_0^5(t^2-t)\ dt=\left[\frac13t^3-\frac12t^2\right]_0^5=\frac13(5)^3-\frac12(5)^2-0=\frac{175}6$$

Answer 3

distance:$$d=\int_0^5|v|\ dt=\int_0^5|t^2-t|\ dt=\int_0^1(t-t^2)\ dt+\int_1^5(t^2-t)\ dt=\dots$$

Answer 4

For Distance \[\int\limits_{0}^{5}\left| t^2-t \right|dt\]
Then since it's always positive I don't need to split it up, correct?
I then evaluated it at 5 and 0 and subtracted and got 125/3 - 25/2 = 175/6m for the distance

Answer 5

you think \(t^2-t>0\) for \(0<t<1\)?

Answer 6

And now just realized I got the same answer for displacement and distance. Is that correct?

Answer 7

no, it's not correct

Answer 8

\(t^2-t<0\) for \(0<t<1\)... see for yourself: \((1/2)^2-1/2=1/4-1/2=-1/4<0\)

Answer 9

distance continued: $$\begin{align*}\int_0^1(t-t^2)\ dt+\int_1^5(t^2-t)\ dt&=\left[\frac12t^2-\frac13t^3\right]_0^1+\left[\frac13t^3-\frac12t^2\right]_1^5\\&=\frac12-\frac13+\frac{125}3-\frac{25}2-\frac13+\frac12\\&=1-\frac23+\frac{175}6\\&=\frac26+\frac{175}6\\&=\frac{177}6\\&=\frac{59}2\end{align*}$$

Answer 10

as you figured out already, \(v(t)<0\) at first which means our particle moves backwards a little before moving forwards eventually: