a particle moves along the x-axis so that its velocity at any time t is given by v(t)=3t^2-18t+24 and its position is given by x(t)=t^3-9t^2+24t+4.
A. For what values of t is the particle at rest?
B. Find the total distance traveled by the particle from T=1 to T=3.
Hint: the particle is moving to the left when v(t)<0 and to the right when v(t)>0.

- anonymous

- katieb

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- anonymous

a) v=0 for rest so solve the quadratic equation of v(t)
b)sketch v(t) for t=1 to v=3 the combination of the area under/over the graph between the x axis is the distance traveled, you may have to do two integrations for this

- anonymous

I don't understand how to get these answers. I have an answer key, I just don't understand velocity at all. I am in precalculus and since it is the end of the year, my teacher is giving us a calculus packet so we can start learning. She has been gone every day since we've gotten it and she will be back tomorrow. I don't understand this velocity stuff at all.

- Loser66

@Narses for partb) why don't we apply directly to x(t) ?

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- anonymous

x(t) may be positive and negative so simply doing x(3) - x(1) may not work

- anonymous

I DON'T GET THISSSS

- anonymous

that would be the magnitude of the distance between the two points not the distance traveled, I was giving a more general case

- Loser66

Nope, not that Narses. the distance can get by absolute value
@bandrockstar wait, after discussing, we will give you the correct instruction.

- anonymous

the total distance traveled may not be the distance between x(1) and x(3)

- anonymous

Okay, thank you.

- Loser66

Ok, guide her steps, please

- anonymous

velocity as a function of time is what you given is describes the speed and which direction, so part a says when is it stationary, that is when the velocity (or speed) is equal to 0 such 3t^2-18t+24 =0

- anonymous

Okay. I got that much now.

- anonymous

so the first step is to set v(t) equal to zero?

- anonymous

Indeed

- anonymous

Okay..

- anonymous

What am I trying to get out of setting it to zero? Like do I basically un-FOIL it?

- anonymous

basically yes use whatever method you know to solve quadratic equations

- anonymous

erm alright. Thanks hold on :)

- anonymous

by setting it to zero

- anonymous

by setting to zero you are telling the equation that the particle is at rest

- anonymous

Okay, That i can actually understand. So what is it after you equal it to zero

- anonymous

so for for part a you have 3t^2-18t+24 =0
that gives t^2-6t+8=0 by dividing both sides by 3 ( I guess you can solve is easier now?)

- anonymous

I don't understand what you put in parenthesis.

- anonymous

I said that you may be able to solve the equation now that we divided by 3?

- anonymous

Yes. is it 3(t-2)(t-4) ?

- anonymous

yes, there is no need for the three as the other side is equal to 0 as 0/3=0

- anonymous

Okay.. so where in the equation (t-2)(t-4) do I get my answer? What am I taking from it? The zeros?

- anonymous

if (t-2)(t-4)=0 imagine a times b =0 how do we get that?
what is a or b?

- anonymous

Like the zeros being 2 and 4? I don't understand what I'm trying to get from this equation

- anonymous

the solution is that either A or B HAASSS to equal 0 so, t-2=0 or t-4 =0 such t=2 and 4

- anonymous

Okay, that's what I was thinking.

- anonymous

So t=2 and t=4 are the values for the particles at rest because thats the values where t=0. Correct?

- anonymous

yes but a typo on your end is you want v=0 at the end

- anonymous

That's what I meant :P How do I do B?

- anonymous

so you have an eqaution for the distance from 0 in terms of t, how would you approach it? any ideas?

- anonymous

Would i be plugging in 1 and 3 somewhere? But I have to use the second equation? I'm just not sure how to do this stuff AT ALL. I have absolutely NO background on it. Sorry :/

- anonymous

its ok, when I first had answered I thought you had basic calculus but no problem, from part a you know that its stationary at t=2 and 4 so in the time period 1->3 the particle must change direction

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- anonymous

Okay. Got it.

- anonymous

as in the diagram shown it 'goes back on itself' so by simply finding the distance between point a and b where t =1 and 3 respectivly the particle will infact travelled further

- anonymous

Yes.

- anonymous

How do we find the distance? Will the second equation be used?

- anonymous

yes, so between t=1 and 2 the particle will be traveling in the same direction so we can take the distance between the two points and do the same with t= 2 and 3

- anonymous

Okay. How do you find the distance between the two?

- anonymous

I need to go soon but work out the difference of x(1) and x(2) and add to the difference of x(2) and x(3) that is the answer,

- anonymous

x(1)= put 1 into to t of x(t)

- anonymous

Thank you so much. I appreciate it very very very much :)

- anonymous

I am not getting the same answer as the key. I get -4 for the difference of x(1) and x(2) and I get 2 for the difference of x(2) and x(3). I am getting 2 when I add them together. The answer on the key is 6 units. I don't understand what I'm doing wrong and I've checked my arithmetic twice.

- anonymous

Uh.. Thank you..

- anonymous

Since you have the velocity at time v (t) = 3t ^ 2-18t +24, then the particle will be at rest when v (t) = 0, that is, you have to find the values of time t for which 0 = 3t ^ 2-18t +24, so you have to solve the quadratic equation. So 0 = 3 (t ^ 2-6t +8) Therefore 0 = (t-4) (t-2), so that t = 4 and t = 2. Thus at t = 2 and t = 4 the particle will be at rest.

- anonymous

ok

- anonymous

I already knew that.

- anonymous

The answer is 6 units.

- anonymous

No, it isn't.

- anonymous

absolute valor (-2)+absolute Valor(4) =6 It is not a vector (Start Point-Final Point), but a scalar, sum of paths.

- anonymous

v (t) = 3t-18t +24 ^ 2 = 3 (t-4) (t-2) is> zero when both (t-4) and (t-2) are positvos or when negative. And t-4> 0 when t> 4 and t-2> 0 when t> 2, so that both are positive parpentesis if t> 4 or both parentheses are negative if t <2, and both are negative paréntsis] 2 , 4 [.
Has to calculate the distance of 1 to 2, d1, 2 = x (2)-x (1), this being the distance to the right (positive sign) and then the distance of 2 to 3, d2, 3 = x (3)-x (2), this being the distance reccorrida left (negative sign), Suma refusal distances more positive one and that is the total distance traveled.
x (2) = 2 ^ 3-9 * 2 ^ 2 +24 * 2 +4 = 24, x (1) = 1 ^ 3-9 * 1 ^ 2 +24 * 1 +4 = 20
d1, 2 = x (2)-x (1) = 24-20 = 4 That is d1, 2 = 4 The distance to the right
x (3) = 3 ^ 3-9 * 3 ^ 2 +24 * 3 +4 = 22 d2, 3 = x (3)-x (2) = 22-24 = -2 The distance to the left from the start point
So the displacement vector is Right + Left = 4-2 = 2 as right shift vector, but if you want the total distance not as vector but a scalar, D=4+2=6

- anonymous

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