# Total distance Issue

• February 13th 2013, 03:18 PM
minneola24
Total distance Issue
Attachment 27004

A bit stuck on this one.

The original equation is x = 1/3t^3 etc etc

Do I set X equal to Y and t is pretty much X?

And do I enter it in on my graphic calculator as such?

A bit confused on the t=3 question too. Total distance isnt the point of (t,y) is it?
• February 13th 2013, 04:57 PM
Soroban
Re: Total distance Issue
Hello, minneola24!

Quote:

$\text{A particle moves along the }x\text{-axis in such a way that}$
$\text{its position at time }t\text{ is given by: }\:x(t)\:=\:\tfrac{1}{3}t^3 - 3t^2 + 8t$

$\text{When }t = 3\text{, what is the total distance the particle has traveled?}$

Solve $v \,=\,\tfrac{dx}{dt}\,=\,0$

. . $t^2 - 6t + 8 \:=\:0 \quad\Rightarrow\quad (x-2)(x-4) \:=\:0 \quad\Rightarrow\quad x \:=\:2,\,4$

Hence, at $t = 2$ and $t=4$, the particle stops and changes direction.

$\text{At }t = 0\!:\;x(0) \:=\:\tfrac{1}{3}(0^3) - 3(0^2) + 8(0) \:=\:0$

$\text{At }t = 2\!:\;x(2) \:=\:\tfrac{1}{3}(2^3) - 3(2^2) + 8(2) \:=\:\tfrac{20}{3}$
. . The particle has moved $6\tfrac{2}{3}$ units to the right.

$\text{At }t = 3\!:\;x(3) \:=\:\tfrac{1}{3}(3^3) - 3(3^2) + 8(3) \:=\:6$
. . The particle has moved $\tfrac{2}{3}$ units to the left.

The particle has traveled: $6\tfrac{2}{3} + \tfrac{2}{3} \:=\:7\tfrac{1}{3}$ units.