1. ## Velocity, Displacement, Distance

The velocity function is v(t)= -t^2 + 6t -8 for a particle moving along a line. Find the displacement and the distance travelled by the particle during the time interval [-1, 5].

2. Hi, My Little Pony

displacement = integral from -1 to 5 (-t^2 + 6t -8) = -18

distance = Sqrt[ ((-5^2+6*5-8) - (1^2-6*1-8))^2) ] = Sqrt[((-3)-(-12))^2]=Sqrt[12^2] = 12 (I'm not sure about this but it looks correctly)

p.s. I'm new on this forum and I don't now how to use math symbols , can anyone help me ?

3. Hello, My Little Pony!

The velocity function is: $v(t)\:= \:-t^2 + 6t -8$ for a particle moving along a line.
Find the displacement and the distance travelled on the time interval [-1, 5].
We have: . $v(t) \:=\:-t^2 + 6t - 8$

The position function is: . $s(t) \:=\:-\tfrac{1}{3}t^3 + 3t^2 - 8t + C$

Assume that $s(0) = 0$, then: $C = 0\quad\Rightarrow\quad s(t) \:=\:-\tfrac{1}{3}t^3 + 3t^2 - 8t$

$s(\text{-}1) \:=\:-\tfrac{1}{3}(\text{-}1)^3 + 3(\text{-}1)^2 - 8(\text{-}1) \:=\:\frac{34}{3}$

$s(5) \:=\:\tfrac{1}{3}(5^3) + 3(5^2) - 8(5) \:=\:-\frac{20}{3}$

The displacement is: . $\left(-\frac{20}{3}\right) - \left(\frac{34}{3}\right) \:=\:-\frac{54}{3} \:=\:-18$

The particle is displaced 18 units to the left.

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Solve $v(t) = 0$

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

. . . This is where the particle stops ... and changes directions.

$s(\text{-}1) \:=\:-\tfrac{1}{3}(\text{-}1) + 3(1) -8(\text{-}1) \:=\:\frac{34}{3}$
$s(2) \:=\:-\tfrac{1}{3}(8) + 3(2^2) - 8(2) \:=\:-\frac{20}{3}$
In the first two seconds, its displacement is: . $-\frac{20}{3} - \frac{34}{3} \:=\:-18$
It has moved $18$ units to the left.

$s(2) \:=\:-\frac{20}{3}$
$s(4) \:=\:-\tfrac{1}{3}(64) + 3(16) - 8(4) \:=\:-\frac{16}{3}$
In the next two seconds, its displacement is: . $-\frac{16}{3} -\left(-\frac{20}{3}\right) \:=\:\frac{4}{3}$
It has moved $\frac{4}{3}$ units to the right.

$s(4) = -\frac{16}{3}$
$s(5) \:=\:-\frac{20}{3}$
In the last second, its displacement is: . $\left(-\frac{20}{3}\right) - \left(-\frac{16}{3}\right) \:=\:-\frac{4}{3}$
It has moved $\frac{4}{3}$ units to the left.

It has moved a total distance of: . $18 + \frac{4}{3} + \frac{4}{3} \:=\:\frac{62}{3} \:=\:20\tfrac{2}{3}$ units.