anonymous
  • anonymous
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.
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • 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
  • 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
  • Loser66
@Narses for partb) why don't we apply directly to x(t) ?

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anonymous
  • anonymous
x(t) may be positive and negative so simply doing x(3) - x(1) may not work
anonymous
  • anonymous
I DON'T GET THISSSS
anonymous
  • 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
  • Loser66
Nope, not that Narses. the distance can get by absolute value @bandrockstar wait, after discussing, we will give you the correct instruction.
anonymous
  • anonymous
the total distance traveled may not be the distance between x(1) and x(3)
anonymous
  • anonymous
Okay, thank you.
Loser66
  • Loser66
Ok, guide her steps, please
anonymous
  • 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
  • anonymous
Okay. I got that much now.
anonymous
  • anonymous
so the first step is to set v(t) equal to zero?
anonymous
  • anonymous
Indeed
anonymous
  • anonymous
Okay..
anonymous
  • anonymous
What am I trying to get out of setting it to zero? Like do I basically un-FOIL it?
anonymous
  • anonymous
basically yes use whatever method you know to solve quadratic equations
anonymous
  • anonymous
erm alright. Thanks hold on :)
anonymous
  • anonymous
by setting it to zero
anonymous
  • anonymous
by setting to zero you are telling the equation that the particle is at rest
anonymous
  • anonymous
Okay, That i can actually understand. So what is it after you equal it to zero
anonymous
  • 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
  • anonymous
I don't understand what you put in parenthesis.
anonymous
  • anonymous
I said that you may be able to solve the equation now that we divided by 3?
anonymous
  • anonymous
Yes. is it 3(t-2)(t-4) ?
anonymous
  • anonymous
yes, there is no need for the three as the other side is equal to 0 as 0/3=0
anonymous
  • 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
  • anonymous
if (t-2)(t-4)=0 imagine a times b =0 how do we get that? what is a or b?
anonymous
  • anonymous
Like the zeros being 2 and 4? I don't understand what I'm trying to get from this equation
anonymous
  • 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
  • anonymous
Okay, that's what I was thinking.
anonymous
  • anonymous
So t=2 and t=4 are the values for the particles at rest because thats the values where t=0. Correct?
anonymous
  • anonymous
yes but a typo on your end is you want v=0 at the end
anonymous
  • anonymous
That's what I meant :P How do I do B?
anonymous
  • anonymous
so you have an eqaution for the distance from 0 in terms of t, how would you approach it? any ideas?
anonymous
  • 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
  • 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
anonymous
  • anonymous
Okay. Got it.
anonymous
  • 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
  • anonymous
Yes.
anonymous
  • anonymous
How do we find the distance? Will the second equation be used?
anonymous
  • 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
  • anonymous
Okay. How do you find the distance between the two?
anonymous
  • 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
  • anonymous
x(1)= put 1 into to t of x(t)
anonymous
  • anonymous
Thank you so much. I appreciate it very very very much :)
anonymous
  • 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
  • anonymous
Uh.. Thank you..
anonymous
  • 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
  • anonymous
ok
anonymous
  • anonymous
I already knew that.
anonymous
  • anonymous
The answer is 6 units.
anonymous
  • anonymous
No, it isn't.
anonymous
  • anonymous
absolute valor (-2)+absolute Valor(4) =6 It is not a vector (Start Point-Final Point), but a scalar, sum of paths.
anonymous
  • 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
  • anonymous

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