How do you find values of t in which the speed of the particle is increasing if he position of a particle moving along a line is given by #s(t)=2t^3-24t^2+90t+7# for #t≥0#?

1 Answer
Feb 17, 2015

If #s(t) = 2t^3 − 24t^2 + 90t + 7#
The speed (at time t) is given by
#s'(t) = 6t^2 -48t +90#

Acceleration is given by the second derivative
# s''(t) = 12 t - 48#

Critical point for acceleration (when the speed changes from decreasing to increasing) occurs when
#s''(t) = 12t - 48 = 0#

That is when #t = 4#

This might be easier to see if we consider a table of time(t) and speed at time t:
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We can see that the speed actually decreases until somewhere between #t=3# and #t=5# when it starts to increase again.