$v(t) = 3t − 8, 0 \leq t \leq 5$
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.
[Math] Find the distance traveled by the particle during the given time interval.
calculus
Best Answer
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 $\frac{113}6$.