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[Lizzie]

A particle moves along a line with velocity function v(t)=t²-t. Find the distance traveled by the particle during the time interval [0,2].

Just a push, please . I am not asking for the answer. Thank you!

 

[pka]

If x(t) is the position function, x’(t)=v(t).
But be careful! The total distance traveled is the definite integral of |v(t)|.

 

[soroban]

Hello, Lizzie!

It's even trickier than pka suggested . . .

A particle moves along a line with velocity function: \(\displaystyle v(t)\:=\:t^2\,-\,t\)
Find the distance traveled by the particle during the time interval [0,2].

First, set \(\displaystyle v(t)\,=\,0:\;\;t^2\,-\,t\:=\:0\;\;\Rightarrow\;\;t(t\,-\,1)\:=\:0\;\;\Rightarrow\;\;t\,=\,0,\,1\)

Why do that? .When \(\displaystyle v\,=\,0\), there is usually a change in direction.


Integrate \(\displaystyle v(t)\) to get the position function:

. . \(\displaystyle s(t)\:=\:\int(t^2\,-\,t)\,dt\;=\;\frac{1}{3}t^3\,-\,\frac{1}{2}t^2\,+\,C\)


At \(\displaystyle t=0:\;s(0)\:=\:\frac{1}{3}(0^3) - \frac{1}{2}(0^2)\,+\,C\:=\:C\)

At \(\displaystyle t=1:\;s(1)\:=\:\frac{1}{3}(1^3)\,-\,\frac{1}{2}(1^2)\,+\,C\:=\:C\,-\,\frac{1}{6}\)
. . It moved \(\displaystyle \frac{1}{6}\) unit to the left.

At \(\displaystyle t=2:\;S(2)\:=\:\frac{1}{3}(2^3)\,-\,\frac{1}{2}(2^2)\,+\,C\:=\:C\,+\,\frac{2}{3}\)
. . It moved: \(\displaystyle (C\,+\,\frac{2}{3})\,-\,(C\,-\,\frac{1}{6})\:=\:\frac{5}{6}\) units to the right.


Distance travelled: \(\displaystyle \;\frac{1}{6}\,+\,\frac{5}{6}\:=\:1\) unit.

 

[pka]

Tricky?
\(\displaystyle \L
\int_0^2 {\left| {t^2 - t} \right|dt} = \int_0^1 {( - t^2 + t)dt} + \int_1^2 {(t^2 - t)dt} = 1\)

 

[Lizzie]

Thank you both for the explanations and the formulas. I really appreciate it.