# hectictar

 Username hectictar Score 9041 Membership Stats Questions 10 Answers 2882

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hectictar  May 29, 2020
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hectictar  Sep 9, 2018
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### Decomposing a Vector

hectictar  Apr 25, 2018
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### Melody's Birthday!!!

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hectictar  Apr 15, 2018
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hectictar  Sep 20, 2017
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hectictar  Apr 25, 2017
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hectictar  Mar 29, 2017
#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.)

Apr 4, 2020
#2
+1

(All except CPhill's signature   )

Guest, if you find the answer on another page, just provide a link to that page. Do not copy the answer as if it were your own. That is plagiarism and it is wrong.

Apr 4, 2020
#1
+2

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 :

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 Apr 4, 2020
#2
+1

2 TiB   =   2 Tebibytes   =   2 * 240  bytes   =   2 * 240 * 8  bits   =   2 * 240 * 23  bits   =   244  bits

There are  244  bits total. We want to know how many  64-bit words  will  "fit into"  that.

number of 64-bit words that can be held  =  \(\dfrac{2^{44}}{64}\ =\ \dfrac{2^{44}}{2^{6}}\ =\ 2^{44-6}\ =\ 2^{38}\)  .

Mar 22, 2020