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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.
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|
Do you know how to get from the position equation to the velocity equation?
take the derivative?
correct. Take the derivative of the position equation and you''ll get the velocity equation.
But we want to know what t is when the V(t)=0
so set the derivative = 0
I got t=3 and t=5
We know that the position \(s(t) = t^3 – 12t^2 + 45t + 4 \) and we also know that \( s′(t) = v(t) \) So, \(s′(t) = v(t) =3t^2-24t+45\)
What does t=3 and t=5 represent?
when the velocity is 0
t=3 and t=5 is the seconds when the velocity is zero
but it's asking about the position and acceleration
How do you find the distance when t=3 seconds and t=5 seconds?
Yes, i know.
the distance formula?
The position equation tells us how far the object went in t=3 and t=5, hence the word position.
so plug those into the position equation?
\(s(t) = t^3 – 12t^2 + 45t + 4\) 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.
so I'm goingto have 2 answers for the position and acceleration?
can you havea negative acceleration? I got 6 and -6
and 54 and 58 for the position
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.
Yes, you are right about the position
ok so thats what i write down?
When t=3, s(t)=54. When t=5, s(t)=58
That's for the position.
But, we don't know what the acceleration is when t=3 and t=5.
You are not done with the question yet.
so i plug in t=3 and t=5 into the second derivative?
i got 6 and -6
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|
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|
ok thanks. i have another question. ill close this then tag you in it ok?