OBJECT MOVES

Ryan Rigdon

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The problem

An object moves along a horizontal coordinate line in such a way that its position at time t is specified by

s = t^3 - 6t^2 - 15t - 3. Here s is measured in centimeters and t in seconds. When is the slowing down; that is,

when is its speed decreasing?


My work.

v = ds/dt = 3t^2 - 12t - 15 = 3(t^2 - 4t - 5) = 3(t-5)(t+1)

the velocity is t = 5, -1 when v=0

the velocity is positive v>0 when (t-5)(t+1)>0 , interval notation (-infinity , -1) U (5, infinity)

Our question was when is the speed decreasing?

v<0 (t-5)(t+1)<0


The objects speed is decreasing during the interval (5 , -1).
 
i did my work over again with an example that i found that was kind of like mine and got the following

v(t) = 3t^2 - 12t - 15

d/dt absolute brackets I 3t^2 - 12t - 15 I = (I 3t^2 - 12t - 15 I)/(3t^2 - 12t - 15) * (6t-12)

= (3 I (t - 5)(t + 1) I)/(3(t-5)(t+1)) * (6t-12) = (I (t -5)(t + 1) I*(6t-12))/((t-5)(t+1)) < 0

t < -1, 1 < t < 5 ; (-infinity , -1) U (1,5)
 
\(\displaystyle Here \ is \ the \ graph \ of \ s(t), \ and \ v(t), \ its \ deriviative \ (black).\)

\(\displaystyle What \ does \ it \ tell \ you?\)

[attachment=0:3bvbm7tl]aaa.jpg[/attachment:3bvbm7tl]
 

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its telling me that the object is decreasing in speed on the interval (-infinity,-1) U (2,5)
 
Ryan Rigdon said:
The problem

An object moves along a horizontal coordinate line in such a way that its position at time t is specified by

s = t^3 - 6t^2 - 15t - 3. Here s is measured in centimeters and t in seconds. When is the slowing down; that is,

when is its speed decreasing?


My work.

v = ds/dt = 3t^2 - 12t - 15 = 3(t^2 - 4t - 5) = 3(t-5)(t+1)

Using a graphing calculator - plot a v-t diagram.

This is a parabola - does it have a maximum or minimum (vertex)? where?

Now think how does the "velocity" behave before/after the vertex?

Keeping in mind that speed indicates the magnitude of velocity - your "final interpretation" ? [(-infinity , -1) U (2,5)] is correct.


the velocity is t = 5, -1 when v=0

the velocity is positive v>0 when (t-5)(t+1)>0 , interval notation (-infinity , -1) U (5, infinity)

Our question was when is the speed decreasing?

v<0 (t-5)(t+1)<0


The objects speed is decreasing during the interval (5 , -1).
 
\(\displaystyle To \ expand \ on \ Subhotosh \ Khan's \ above \ comment, \ observed \ the \ following \ graph,\)

\(\displaystyle where \ the \ green \ parabola \ = \ velocity \ and \ the \ black \ line \ = \ acceleration.\)

\(\displaystyle Now, \ the \ speed \ of \ a \ particle \ is \ increasing \ if \ its \ velocity \ and \ acceleration \ have \ the \ same\)

\(\displaystyle sign, \ and \ the \ speed \ is \ decreasing \ if \ they \ have \ opposite \ signs.\)

\(\displaystyle Hence: \ Particle \ increasing \ on \ (-1,2)U(5,\infty) \ and \ decreasing \ on \ (-\infty,-1)U(2,5)\)

\(\displaystyle In \ physicists's \ terms, \ the \ particle \ starts \ at \ negative \ \infty, \ close \ to \ the \ speed \ of\)

\(\displaystyle light, \ slows \ down \ to \ 0 \ at \ t \ = \ -1, \ then \ speeds \ up \ to \ the \ inflection \ point \ at \ t \ = \ 2,\)

\(\displaystyle slows \ down \ again \ to \ 0 \ at \ t \ = \ 5, \ then \ speeds \ up \ again \ to \ \infty, \ reaching \ close \ to \ the\)

\(\displaystyle speed \ of \ light.\)

\(\displaystyle Note: \ According \ to \ Einstein, \ nothing \ can \ exceed \ or \ equal \ the \ speed \ of \ light. \ Can \ come \ close,\)

\(\displaystyle but \ no \ cigar.\)

\(\displaystyle Additional \ note, \ a \ minor \ digression \ to \ the \ problem \ at \ hand.\)

\(\displaystyle Einstein's \ theory \ on \ the \ speed \ of \ light \ is \ just \ that \ a \ theory, \ however \ if \ true, \ then \ we \ as \ a\)

\(\displaystyle species \ are \ doomed \ to \ live \ and \ die \ and \ eventually \ become \ extinct\)

\(\displaystyle as \ we \ are \ trapped \ in \ the \ "proverbial \ trickbag", \ our \ present \ solar \ system, \ as \ the\)

\(\displaystyle nearest \ star, \ other \ than \ our \ sun \ is \ about \ 4 \ light \ years \ hence. \ Worm \ holes \ anyone?\)

[attachment=0:j7chjl0u]bbb.jpg[/attachment:j7chjl0u]
 

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