Expert: Sherman D. - 11/7/2005

Question
At time t, the position of a body moving along the s-axis is s=t^3-6t^2+9t m.

1) find the body's acceleration each time the velocity is 0.
2) find the body's speed each time the acceleration is 0.
3) find the total distance traveled by the body from t=0 to t=2.

Answer
Sorry it is taking me so long, but before i can start helping you, i need to first get some help from someone else, so that i won't need any help for other questions that you may have.

Here is what i can give you

1.)
s = t^3 - 6t^2 + 9t
s' = 3t^(3 - 1) - 12t^(2 - 1) + 9t^(1 - 1)
s' = 3t^2 - 12t + 9
v = 3t^2 - 12t + 9

0 = 3t^2 - 12t + 9
0 = 3(t^2 - 4t + 3)
0 = t^2 - 4t + 3
0 = (t - 3)(t - 1)
t = 1 or 3

v = 3t^2 - 12t + 9
v' = 3t^2 - 12t
v' = 6t^(2 - 1) - 12t^(1 - 1)
v' = 6t - 12
a = 6t - 12

a = 6(1) - 12
a = 6 - 12
a = -6
or
a = 6(3) - 12
a = 18 - 12
a = 6

ANS : Accleration = -6m/s^2 and 6m/s^2

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2.)
a = 6t - 12
0 = 6t - 12
12 = 6t
t = 2

v = 3t^2 - 12t + 9
v = 3(2)^2 - 12(2) + 9
v = 3(4) - 24 + 9
v = 12 - 24 + 9
v = -12 + 9
v = -3 m/s

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3.)
s = t^3 - 6t^2 + 9t
s = 2^3 - 6(2)^2 + 9(2)
s = 8 - 6(4) + 18
s = 8 - 24 + 18
s = -16 + 18
s = 2 meters

but i am not completely certain.