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!

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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!

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Answer 1 · 2018-05-08T21:41:03

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

Explanation:

Started with
$s(t) = 2t^3 - 24t +1$

Differentiated it
$s'(t) = 6t^2 - 24$

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

$color(red)(s'(t))$ --the differentiated equation--equals to $color(red)m$ (slope of gradient) so...You get: $m = s'(t)$

You then put the values that you have into $m = s'(t)$
Your values:
$color(blue)(m = 0)$
$color(blue)(s'(t) = 6t^2 - 24)$

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

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

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

$24/6 = t^2$
$4 = t^2$

$sqrt4 = t$
$2 = t$

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

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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!

1 Answer

Explanation:

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