A particle moves along a straight line and its position at time t is given by: 2t^3-18t^2+48t, t>=0?

Question

where s is measured in feet and t in seconds.
Use interval notation to indicate the time interval or union of time intervals when the particle is moving forward and backward.
Forward=
Backward=
Use interval notation to indicate the time interval(s) when the particle is speeding up and slowing down.
Speeding up=
Slowing down=
Thanks so much!

Answer 1

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velocity = derivative of position
v(t) = s'(t)
s(t) = 2t^3 - 18t^2 + 48t
s'(t) = 6t^2 - 36t + 48 = velocity

particle is moving forward when v(t) > 0 and backwards when v(t) < 0 find the zeros of the velocity function and then test points between the zeros (when particle has stopped to reverse direction) to find whether it is moving forwards or backwards.

v(t) = 6t^2 - 36t + 48 = 6[t^2 - 6t + 8] = 6(t-4)(t-2) v(t) = 0 at t = 4 and t = 2
point t = 1, (v1) = 6 - 36 + 48 > 0 so moving forwards
point t = 3, v(3) = 54 - 108 + 48 < 0 so moving backwards
point t = 5, v(5) = 150 - 180 + 48 > 0 so moving forwards

forward: between 0 and 2 and points greater than 4
backward: between 2 and 4
(I can't remember interval notation)

acceleration is the derivative of the velocity function
a(t) = v'(t)
v(t) = 6t^2 - 36t + 48
v'(t) = 12t - 36
a(t) = 12t - 36
find zero
12t-36 = 0, t = 3
plug in points before and after 3
point t = 0: 12(0)-36 <0 so slowing down
point t = 4: 48 - 36 > 0 so speeding up

Speeding up: points greater than t = 3
Slowing down: points between 0 and 3