$v(t) = 3t − 8, 0 \leq t \leq 5$

I keep getting $143/6$ as my answer but apparently it's not correct. Can someone please help me out? I'm about to pull my hair out working on this problem.

  • 1
    $\begingroup$ Hello! Could you show your work please? You will be more likely to get help. Thanks! $\endgroup$ – user130512 Apr 23 '14 at 19:12
  • $\begingroup$ Give me a minute. $\endgroup$ – James Apr 23 '14 at 19:13
  • $\begingroup$ What is $v(t)$? $\endgroup$ – Mhenni Benghorbal Apr 23 '14 at 19:17
  • $\begingroup$ v=velocity, I assume $\endgroup$ – Hagen von Eitzen Apr 23 '14 at 19:17
  • $\begingroup$ You need to add more information into your question. I can guess that $v(t)$ is the instantaneous velocity too but your question should be complete. $\endgroup$ – Mhenni Benghorbal Apr 23 '14 at 19:21

We have $$\int_0^5(3t-8)\,\mathrm dt =\left.\frac32t^2-8t\right|_0^5=-\frac52$$ so the particle ends up $\frac52$ units "to the left" of the starting position. Thus the total net distance travelled is $\frac52$ units, or $-\frac52$ if you take the displacement with sign.

One could interprete "distance travelled" differently, insofar as the particle first moves to the left (until $t=\frac 83$) and then to the right, i.e. it moves from $x(0)=0$ via $$ x(8/3)=\int_0^{\frac83}(3t-8)\,\mathrm dt =\left.\frac32t^2-8t\right|_0^{\frac83}=-\frac{32}3$$ to the end point at $x(5)=-\frac52$. So the particle has travelled $\frac{32}3$ units in the first part and $\left|-\frac52-(-\frac{32}3)\right|=\frac{49}6$ in the second part, hence a total distance of $\frac{113}6$.


In this example, $v$ is the velocity. We have $v(t) = 3t-8$ and it's important to notice that $v < 0$ when $t<\frac{8}{3}$, $v=0$ when $t=\frac{8}{3}$ and $v>0$ when $t>\frac{8}{3}$.

The "story" of the particle is that it moves to the left for all $0 \le t <\frac{8}{3}$, it stops for an instant when $t=\frac{8}{3}$, and then it starts to move to the right for all $t>\frac{8}{3}$.

To find the distance (and not the displacemenet), we can integrate the velocity.

For the motion to the left we calculate $$\int_0^{8/3} 3t -8~\mathrm{d}t = \left[ \frac{3}{2}t^2-8t\right]_0^{8/3}=-\frac{32}{3}$$ The negative sign tells us it is a distance traveled to the left.

Next we find the distance traveled to the right

$$\int_{8/3}^5 3t-8 ~ \mathrm{d}t = \left[\frac{3}{2}t^2-8t\right]_{8/3}^5 = \frac{49}{6}$$

Having moved $\frac{32}{3}$ to the left and then $\frac{49}{6}$ to the right, our total distance is

$$\frac{32}{3} + \frac{49}{6} = \frac{113}{6} = 18.8\overline{3}$$


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