anonymous
  • anonymous
Calc Help
Mathematics
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chestercat
  • chestercat
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anonymous
  • anonymous
The position function of a particle in rectilinear motion is given by s(t) = t3 – 12t2 + 45t + 4 for t ≥ 0.Find the position and acceleration of the particle at the instant the when the particle reverses direction. Include units in your answer.
Zale101
  • Zale101
Your question asks "Find the position and acceleration of the particle at the instant the when the particle reverses direction. " When the particle reverses direction, we know the velocity equals zero. |dw:1450392161868:dw|
Zale101
  • Zale101
Do you know how to get from the position equation to the velocity equation?

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anonymous
  • anonymous
take the derivative?
Zale101
  • Zale101
correct. Take the derivative of the position equation and you''ll get the velocity equation.
anonymous
  • anonymous
3t^2-24t+45
Zale101
  • Zale101
But we want to know what t is when the V(t)=0
anonymous
  • anonymous
so set the derivative = 0
anonymous
  • anonymous
I got t=3 and t=5
Zale101
  • Zale101
We know that the position \(s(t) = t^3 – 12t^2 + 45t + 4 \) and we also know that \( s′(t) = v(t) \) So, \(s′(t) = v(t) =3t^2-24t+45\)
Zale101
  • Zale101
What does t=3 and t=5 represent?
anonymous
  • anonymous
when the velocity is 0
Zale101
  • Zale101
t=3 and t=5 is the seconds when the velocity is zero
Zale101
  • Zale101
Now|dw:1450392689390:dw|
anonymous
  • anonymous
but it's asking about the position and acceleration
Zale101
  • Zale101
How do you find the distance when t=3 seconds and t=5 seconds?
Zale101
  • Zale101
Yes, i know.
anonymous
  • anonymous
the distance formula?
Zale101
  • Zale101
The position equation tells us how far the object went in t=3 and t=5, hence the word position.
anonymous
  • anonymous
so plug those into the position equation?
Zale101
  • Zale101
\(s(t) = t^3 – 12t^2 + 45t + 4\) this is your position equation Your question says, what is the position and acceleration when t=3 and t=5 Yes, you plug in t=3 and t=5 in the position equation to see how far the object went at that time.
anonymous
  • anonymous
so I'm goingto have 2 answers for the position and acceleration?
Zale101
  • Zale101
Yes.
anonymous
  • anonymous
can you havea negative acceleration? I got 6 and -6
anonymous
  • anonymous
and 54 and 58 for the position
Zale101
  • Zale101
Yes you can have a negative equation. If the acceleration goes in the opposite direction, it is negative. If you throw in object, when it's falling down, you'll get -9.8 m/s^2 by gravity.
Zale101
  • Zale101
Yes, you are right about the position
anonymous
  • anonymous
ok so thats what i write down?
Zale101
  • Zale101
When t=3, s(t)=54. When t=5, s(t)=58
Zale101
  • Zale101
That's for the position.
Zale101
  • Zale101
But, we don't know what the acceleration is when t=3 and t=5.
Zale101
  • Zale101
You are not done with the question yet.
anonymous
  • anonymous
so i plug in t=3 and t=5 into the second derivative?
Zale101
  • Zale101
Yes.
anonymous
  • anonymous
i got 6 and -6
Zale101
  • Zale101
Correct.
Zale101
  • Zale101
The reason why you get two answers for the seconds, position, and acceleration is because your position equation is in 3rd degree. That means the end behavior goes up and down since the degree is an odd number (3). |dw:1450393489449:dw|
Zale101
  • Zale101
If it was an even degree, like the parabola, we would get only one seconds when solving for t after setting v(t)=0|dw:1450393676421:dw|
anonymous
  • anonymous
ok thanks. i have another question. ill close this then tag you in it ok?
Zale101
  • Zale101
|dw:1450393696295:dw|
Zale101
  • Zale101
Okay.

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