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
Vince G.
May 8, 2018

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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