#1**+2 **

s(t) = t^{3} - 6t^{2} + 9t + 1

**a:**

v(t) = s'(t) because velocity is the derivative of position

v(t) = 3t^{2} - 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

hectictar Apr 4, 2020

#3**+3 **

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.)

hectictar Apr 5, 2020