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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.

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

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

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