# A particle moves along a line so that its displacement at time t≥ 0 is given by s(t)=2t^3-24t+1, where s is measured in meters and t is measured in seconds. At what values of t does the particle change direction? Thank you!

May 8, 2018

By using graphs and a bit of differentiation, I got t = 2 and t = -2

#### Explanation:

Started with
$s \left(t\right) = 2 {t}^{3} - 24 t + 1$

Differentiated it
$s ' \left(t\right) = 6 {t}^{2} - 24$

In a graph, a 'turning point' or in other words, change in direction means that m(slope of graph) = 0.

$\textcolor{red}{s ' \left(t\right)}$--the differentiated equation--equals to $\textcolor{red}{m}$(slope of gradient) so...You get: $m = s ' \left(t\right)$

You then put the values that you have into $m = s ' \left(t\right)$
$\textcolor{b l u e}{m = 0}$
$\textcolor{b l u e}{s ' \left(t\right) = 6 {t}^{2} - 24}$

$m = s ' \left(t\right)$ turns to color(red)(0 = 6t^2 - 24

When you solve that equation above, you get...
color(red)(0 = 6t^2 - 24

$0 + 24 = 6 {t}^{2}$
$24 = 6 {t}^{2}$

$\frac{24}{6} = {t}^{2}$
$4 = {t}^{2}$

$\sqrt{4} = t$
$2 = t$

Since with your starting equation...color(red)(s(t) = 2t^3 - 24t +1...is a cubic equation (a cubic equation is the one the looks like an 'N' in a graph), you have 2 turning points.

Therefore, $t = 2$ is not right because that would mean there's only 1 turning point so...you do $t = 2$ or $- 2$

and that's the answer, i think