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Finding velocity of a particle

  1. Sep 18, 2008 #1
    1. The problem statement, all variables and given/known data
    The position of a particle is given by: x = 2 + 3t - 4t^2. Determine the position of the particle when it changes direction. Determine its velocity when it returns to the position it had at t = 0 sec. x is in met. and t is in sec.


    2. Relevant equations
    I calculated the particle's position by derivating f '(x) = 3t - 4t^2 = 0

    3. The attempt at a solution
    f ' (x) = 3t - 4t^2 = 0 = 3 - 8t
    t = 0.375 sec.
    x = 2 + 3(0.375) - 4(0.375)^2
    x = 2.5625 met. position

    My question is: would this be the correct way to find the velocity of the particle(see the problem statement):
    3 - 8t = 0
    3 - 8(0 sec.) = 3 met/sec. ??
     
  2. jcsd
  3. Sep 19, 2008 #2
    It works due to the fact that it takes the particle the same amount of time to go from it's starting point to the point it turns and from the point it turns around to the starting point. However, your answer is still wrong, but only slightly.
     
  4. Sep 19, 2008 #3

    alphysicist

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    Hi science_rules,


    I believe you might have misread the question. They don't want the velocity at [itex]t=0[/itex], they want the velocity when it is at the same position at [itex]t=0[/itex].

    In other words, what is the particle's position at [itex]t=0[/itex]? At what later time [itex]t[/itex] has the particle returned to that same position? Then, what is the velocity at that later time?
     
  5. Sep 19, 2008 #4
    then would it be correct to use this equation to find the velocity?:
    v^2 = vi^2 + 2aS i know S stands for the distance. i convert all numbers to met/sec.
    = (3.0 X 10^6 met/sec)^2 + 2(1.196 X 10^15 met/sec.^2)(1.5 X 10^-2 met)
    = (9 X 10^12 met/sec) + 2(1.794 X 10^13 met/sec)
    = (9 X 10^12) + (3.588 X 10^13)
    = 4.488 X 10^13 meters/sec
     
  6. Sep 19, 2008 #5

    alphysicist

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    I don't see where you are getting these numbers from. Are you accidentally using the numbers from a different problem?

    This equation in a sense applies to this problem, but it would be more of a conceptual approach.

    My last post showed three questions to answer to solve this problem. They were:

    What is the particle's position at t=0?
    At what later time has the particle returned to that same position?
    Then, what is the velocity at that later time?

    You already have the two equations in your original post that you need to answer these three questions, and the last question solves the problem. What do you get?


    EDIT: I saw your PM, and I am only referring to part b of this problem. I believe the first part (finding the position where it turns around) is correct.
     
  7. Sep 19, 2008 #6
    as a matter of fact i accidently used numbers-the 3 X 10^6 etc. from a problem right underneath the problem i'm trying to do in my book. i hate that! rrrghh. lol
     
  8. Sep 19, 2008 #7
    if i'm not mistaken, i think i can use the quadratic equation to solve for t. i assume x is set to 2 meters because that is where i am trying to find at what later time the particle returned to that position?

    so....
    x = 2 + 3t - 4t^2
    2 = 2 + 3t - 4t^2 now i believe i need the quadratic equation to find t.

    ax^2 + bx + c

    delta = b^2 - 4ac = 3^2 - 4(-4)(2) = 41
    now i use the quadratic formula:

    x = -b +/- squrtdelta / 2a

    first x = -3 + squrt 41 / 2(-4) = -0.425 met.
    second x = -3 - squrt 41 / 2(-4) = -9.403 met.

    then i plug one of those into v = 3-8t to get the velocity.
    this is where i am confused again. neither of those numbers seem right:
    v = 3-8(-0.425) = 6.4 met/sec
    v = 3-8(-9.403) = -72.22 met/sec
    none of those are close to 3 met/sec. as it should be, but i think i used the quad. equation correctly. does it have to be near 3 met/sec?
     
  9. Sep 19, 2008 #8
    oops i forgot to say that the later time is t = -0.425 sec. or is it t = -9.403? when (x = 2)
    then i have to check if either of those times equals 2.
     
  10. Sep 19, 2008 #9
    i mean, if when i plug in either of those times, it makes the other side of the equation equal to 2 meters.
     
  11. Sep 19, 2008 #10
    it looks like the t = -0.425 is the closer one, but it still doesnt make the equation equal to 2 meters. darn. im confused now. all this calculating and it seems im not getting anywhere :(
     
  12. Sep 19, 2008 #11
    2 = 2 + 3(-0.425) - 4(-0.425)^2
    = 2 - 1.275 + 0.7225
    = 1.44
     
  13. Sep 19, 2008 #12

    alphysicist

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    This equation is right, but you don't need the quadratic formula to solve this. First subtract 2 from both sides, and you'll get

    [tex]
    0=3t-4t^2
    [/tex]

    With this, you can divide everything by t, and then solve what's left for the t that you are looking for. What do you get? (Once you get it, you can check it by plugging it back into [itex]0=3t-4t^2[/tex] to verify that the right side does become zero with that value of t.) What do you get?
     
  14. Sep 19, 2008 #13
    sigh. silly me. well, atleast i remember how to use the quadratic equation.
    so, then it would be:
    2= 2 + 3t -4t^2 subtract 2 from both sides.
    0 = 3t - 4t^2
    0/t = 3t - 4t^2 / t (zero divided by any number will be zero)
    0 = 3 - 4t subtract the 3
    -3 = -4t

    -3/-4 = 0.75 seconds

    put into v = 3 - 8t

    v = 3 - 8(0.75)
    = 3 - 6
    = -3 meters/sec. woohoooo!!! so it's negative 3. you're right it is a 3 but it's a negative 3. cool
     
  15. Sep 19, 2008 #14
    it's tough in physics, learning when to not complicate things too much, or learning to realize you're missing certain steps you should do in order to solve a problem! i wonder when i'll learn! i hope i can get good enough with physics to help others someday.
     
  16. Sep 19, 2008 #15
    oh check:
    0 = 3t - 4t^2
    0 = 3(0.75) - 4(0.75)^2
    0 = 2.25 - 4(0.5625)
    0 = 2.25 - 2.25 = 0 yay!
     
  17. Sep 19, 2008 #16
    i declare this problem solved :)
     
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