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cgbork

  • one year ago

A particle's position is r(vector) =(ct^2−2dt^3)i^+(2ct^2−dt^3)j^, where c and d are positive constants. Find expressions for times t > 0 when the particle is moving in the x-direction. Express your answer in terms of the variables c and d. can anybody walk me through how to even approach this problem? I am totally lost.

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  1. anonymous
    • one year ago
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    Have you taken calculus by any chance? And If so is this a calculus based class?

  2. anonymous
    • one year ago
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    I can walk you through it but I need to think about how to tailor my approach based on your level.

  3. cgbork
    • one year ago
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    Yes. This is calc based. I believe the derivative needs to be taken to solve this

  4. anonymous
    • one year ago
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    Correct the instantaneous rate of change of position is velocity, and that can be calculated by taking the derivative. This seems function seems to be fairly tame, do you recall the derivative rule for polynomials?

  5. cgbork
    • one year ago
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    not entirely. I am on a site for my class where it deducts 5% every time I guess wrong. I have put in d^2r/dt^2 and it said it didn't depend on dt or r

  6. anonymous
    • one year ago
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    Sorry I didnt see that you had returned hold on...

  7. anonymous
    • one year ago
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    Ok is this the function you gave:

  8. anonymous
    • one year ago
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    \[ r(t)= (c*t^{2} - (2d)*t^{3})i+((2c)*t^{2} - d* t^{3})j \]

  9. anonymous
    • one year ago
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    Note how I made sure not to run the constants together with the independent variable... * denotes typical multiplication but I added it explicitly to make sure this is what you had in the problem statement

  10. anonymous
    • one year ago
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    Are you still there?

  11. cgbork
    • one year ago
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    Internet failed

  12. anonymous
    • one year ago
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    Hello

  13. anonymous
    • one year ago
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    So if a particle is moving in the x-direction what does that mean?

  14. anonymous
    • one year ago
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    It means the x-component of its velocity is non-zero... Please check to make sure you copied the question correctly because I suspect something is missing here.

  15. cgbork
    • one year ago
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    I believe it is correct

  16. cgbork
    • one year ago
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    My hint was this.. Hint 1. You will have to take a derivative with respect to time to get the velocity of the particle.

  17. anonymous
    • one year ago
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    Correct, but what I am getting at is only requiring the x-component of the velocity v_x to be non-zero results in only one point where t>0 and v_x is zero and that is t=c/(3d).... I was expecting an interval or something more than a single point

  18. anonymous
    • one year ago
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    So in other words any time t>0 except fot that point t=c/(3d) the particle is moving in the x direction

  19. cgbork
    • one year ago
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    I already tried t=c/3d and it said I was missing something

  20. cgbork
    • one year ago
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    Specifically, Your answer either contains an incorrect numerical multiplier or is missing one.

  21. anonymous
    • one year ago
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    No at that point there is NO motion in the x-direction. At that point v_x is zero... at all other times besides that point there is some motion in the x-direction

  22. cgbork
    • one year ago
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    Oh crap. That is because I found it when t=0 and it was asking for when it is greater than 0. So how would I find it when it isn't 0, but instead t>0?

  23. anonymous
    • one year ago
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    No that is the answer for when t > 0. When t=0 the answer is just v_x = 0 plain and simple but when t >0 you get c-3dt=0 or t = c/(3d)

  24. anonymous
    • one year ago
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    Like I said I feel like this question is either ill posed or incorrect... If that isn't the answer.... all t =/= c/(3d)

  25. anonymous
    • one year ago
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    Does it mean in the POSITIVE x direction maybe?

  26. cgbork
    • one year ago
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    It was asking for t>0 but told me that c/3d was wrong. This computer program for this question is on crack.

  27. anonymous
    • one year ago
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    Yes because it IS wrong.... The answer is all t NOT equal to that value

  28. cgbork
    • one year ago
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    So I need to make an equation that excludes that point

  29. anonymous
    • one year ago
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    Assuming any motion in the x-direction (either way) is permitted.

  30. anonymous
    • one year ago
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    sure try 0<t<c/(3d) and c/(3d)<t<infinity

  31. cgbork
    • one year ago
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    Part a is asking for the x direction and part b is asking for the y direction. So I'm hoping once I understand how to get part a that part b will be a breeze. Are you sure it is asking for all of that?

  32. anonymous
    • one year ago
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    No Im not sure like I said I feel like this question is missing some information or poorly written or something. I mean have you tried the above answer?

  33. anonymous
    • one year ago
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    Having only a single point seems kind of silly to me... I feel like it should be an interval or something.... Asking only about POSITIVE x-direction or something would make more sense

  34. anonymous
    • one year ago
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    But that is just a guess

  35. cgbork
    • one year ago
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    All of 0<t<c/(3d) and c/(3d)<t<infinity? No, I haven't put that in yet. The whole MasteringPhysics online program is definitely confusing. What would I have to do if it was in positive x direction? I'm going to assume it is because it didn't say anything differently. Thank you for your help btw!

  36. anonymous
    • one year ago
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    if it were in the positive x-direction then the as long as 2ct > 6dt^2 or as long as 2c >6dt or c/(3d) > t

  37. anonymous
    • one year ago
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    So for all times 0 < t < c/(3d)

  38. anonymous
    • one year ago
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    After that the velocity becomes negative (moves in the -x-direction) and stays that ways forever

  39. anonymous
    • one year ago
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    Did you try that?

  40. cgbork
    • one year ago
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    I don't know if I can put it in like that

  41. cgbork
    • one year ago
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    I figured it out. Yay for physics roommate! Thank you for trying!!!

  42. anonymous
    • one year ago
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    So what did the answer end up being?

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