+9479 s(t) = t3 - 6t2 + 9t + 1
a:
v(t) = s'(t) because velocity is the derivative of position
v(t) = 3t2 - 12t + 9 This equation tells us her velocity at any time t . To find her velocity when t = 3 , plug in 3 for t
v(3) = 3(3)2 - 12(3) + 9
v(3) = 0 (This is is meters per second)
b:
a(t) = v'(t) because acceleration is the derivative of velocity
a(t) = 6t - 12 Now we can find what value of t causes a(t) to be 0
6t - 12 = 0 Solve this equation for t
t = 2
So when t = 2 her acceleration is 0 . But we want to know what is her velocity when t = 2 ?
v(2) = 3(2)2 - 12(2) + 9
v(2) = -3 (This is in meters per second)
c:
Alice is at rest when her velocity is zero. Can you find what values of t make v(t) be 0 ?
d:
She's going backwards when her velocity is negative. So after finding what values of t make her velocity 0 , we can test a t value on each side of each zero to find the intervals where the velocity is negative.
e:
Here's a graph of s(t) with points on the graph labeled for t = 1,2,3,4,5,6 :
https://www.desmos.com/calculator/ryywlaotbp
I'm kind of guessing on this one, but here's my guess:
After 0 s , she is 1 m from the starting position, and she moved a total distance of 0 m
After 1 s , she is 5 m from the starting position, and she moved a total distance of 4 m
After 2 s , she is 3 m from the starting position, and she moved a total distance of 6 m
After 3 s , she is 1 m from the starting position, and she moved a total distance of 8 m
After 4 s , she is 5 m from the starting position, and she moved a total distance of 12 m
After 5 s , she is 21 m from the starting position, and she moved a total distance of 28 m
After 6 s , she is 55 m from the starting position, and she moved a total distance of 62 m
Let us know if you have a question or need more help on this 
+9479 At 1 second and at 3 seconds is correct for c .
(Notice how on this graph, when t = 1 and t = 3, the tangent line is totally "flat" with a slope of zero.)
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Alice is moving backwards when the velocity is negative.
Here is a graph of the velocity function for reference: https://www.desmos.com/calculator/dqfc3ssois
(Note that in the graphs, x represents t )
We know that the graph is continuous everywhere,
AND we know that the only place where the graph touches the horizontal axis is at t = 1 and at t = 3
So is the graph above or below the horizontal axis when t < 1 ?
To find that out, we can test any value that is less than 1 . So let's use t = 0
v(0) = 3(0)2 - 12(0) + 9
v(0) = 9
9 is positive, so when t = 0 , the velocity is positive.
And we can conclude that, for ALL values of t less than 1 , the velocity is positive.
v(t) > 0 for all values of t in the interval (-∞, 1)
Now we want to know is the graph above or below the horizontal axis when t is between 1 and 3 ?
To find that out, we can test any value that is between 1 and 3 . So let's use t = 2
v(2) = 3(2)2 - 12(2) + 9
v(2) = -3
-3 is negative, so when t = 2 , the velocity is negative.
And we can conclude that, for ALL values of t between 1 and 3 , the velocity is negative.
v(t) < 0 for all values of t in the interval (1, 3)
Now we want to know is the graph above or below the horizontal axis when t is greater than 3 ?
To find that out, we can test any value that is greater than 3 . So let's use t = 4
v(4) = 3(4)2 - 12(4) + 9
v(4) = 9
9 is positive, so when t = 4 , the velocity is positive.
And we can conclude that, for ALL values of t greater than 3 , the velocity is positive.
v(t) > 0 for all values of t in the interval (3, ∞)
So now we know that Alice is moving backwards when t is in the interval (1, 3)
(Notice how on this graph, when t is in the interval (1, 3), the tangent line is "downhill" with a negative slope.)
