## bandrockstar 2 years ago 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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1. Narses

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

2. bandrockstar

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.

3. Loser66

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

4. Narses

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

5. bandrockstar

I DON'T GET THISSSS

6. Narses

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

7. Loser66

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

8. Narses

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

9. bandrockstar

Okay, thank you.

10. Loser66

Ok, guide her steps, please

11. Narses

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

12. bandrockstar

Okay. I got that much now.

13. bandrockstar

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

14. Narses

Indeed

15. bandrockstar

Okay..

16. bandrockstar

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

17. Narses

basically yes use whatever method you know to solve quadratic equations

18. bandrockstar

erm alright. Thanks hold on :)

19. Narses

by setting it to zero

20. Narses

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

21. bandrockstar

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

22. Narses

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?)

23. bandrockstar

I don't understand what you put in parenthesis.

24. Narses

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

25. bandrockstar

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

26. Narses

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

27. bandrockstar

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

28. Narses

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

29. bandrockstar

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

30. Narses

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

31. bandrockstar

Okay, that's what I was thinking.

32. bandrockstar

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

33. Narses

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

34. bandrockstar

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

35. Narses

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

36. bandrockstar

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 :/

37. Narses

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

38. bandrockstar

Okay. Got it.

39. Narses

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

40. bandrockstar

Yes.

41. bandrockstar

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

42. Narses

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

43. bandrockstar

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

44. Narses

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,

45. Narses

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

46. bandrockstar

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

47. bandrockstar

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.

48. bandrockstar

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.

50. bandrockstar

ok

51. bandrockstar

I already knew that.

52. bandrockstar

The answer is 6 units.

53. bandrockstar

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.