1. ## Calc Word Problem

A particle moves along a line. The particle's position, $s$, in meters at $t$ seconds is modelled by $s(t) = 2t^{3} - 15t^2 + 36t + 40$, where $t \geq 0$.

e) Determine the total distance travelled during the first five seconds

Found the first derivative, used interval chart, found distances for time found and tried to find the total distance for the first 5 seconds:

$0 = 6t^2 - 30t + 36$
$0 = 6(t - 3)(t - 2)$
$t = 2s$ and $t = 3s$

$(0, 2) \rightarrow$ advancing
$(2, 3) \rightarrow$ retreating
$(3, \infty) \rightarrow$ advancing

$s(0) = 40m$
$s(2) = 68m$
$s(3) = 67m$
$s(5) = 95m$

$s(0) = 40m$
$s(3) - s(2) = 1m$
$s(5) - s(3) = 28m$

$40m + 1m + 28m = 69m$

2. Originally Posted by Macleef
A particle moves along a line. The particle's position, $s$, in meters at $t$ seconds is modelled by $s(t) = 2t^{3} - 15t^2 + 36t + 40$, where $t \geq 0$.

e) Determine the total distance travelled during the first five seconds

Found the first derivative, used interval chart, found distances for time found and tried to find the total distance for the first 5 seconds:

$0 = 6t^2 - 30t + 36$
$0 = 6(t - 3)(t - 2)$
$t = 2s$ and $t = 3s$

$(0, 2) \rightarrow$ advancing
$(2, 3) \rightarrow$ retreating
$(3, \infty) \rightarrow$ advancing

$s(0) = 40m$
$s(2) = 68m$
$s(3) = 67m$
$s(5) = 95m$

$s(0) = 40m$
$s(3) - s(2) = 1m$
$s(5) - s(3) = 28m$

$40m + 1m + 28m = 69m$

$s(0)=40m$ is where you started
$s(2)=68m$ is where it turns around
so the first displacement should be $s(2)-s(0)=28$
Then you get $28+1+28=57m$