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Kinematic Equations

  1. Oct 2, 2013 #1
    Hey guys how's it going. My physics teacher gave us a challenge problem for extra credit, me needed the extra credit i decided to come here for help.

    1. The problem statement, all variables and given/known data
    here is the link to the problem itself. http://www.modelingphysics.org/consta/challenge.pdf

    I have actually done number 1, i can't cant seem to figure out how exactly to piece together 2. I have to come up with a algebraic equation to show exactly when the carts will collide.

    Given is:
    Cart A: Starts at a position of 0, velocity of zero, and an unknown a acceleration
    Cart B: Starts at unknown position, velocity of zero, and unknown acceleration.




    2. Relevant equations

    Exact problem is in link. As far as equations go it uses the kinematic equations



    3. The attempt at a solution

    My physics teachers told us to put the variables in a list then plug them into a kinematic equation to solve for a unknown value then.
    Cart A:
    X=0
    V1=0
    A=?
    Vf=?
    t=?
    Cart B:
    X=?
    v1=0
    Vf=?
    t=?
    A=?
    _______________________________________________________________________________
    A= Acceleration
    T= Time
    V1= initial velocity
    X= Position
    VF= Final velocity

    Im only confused as this involves puting them into a equation that has them colliding and that is where im stuck.

    I would gladly appreciate the help. Thanks
     
  2. jcsd
  3. Oct 3, 2013 #2
    When the two carts collide, they are at the same position at the same time. What are the kinematic equations involving time and position?
     
  4. Oct 3, 2013 #3
    I used x=v1+1/2at^2.
    As the equation. How would I go from there?
     
  5. Oct 3, 2013 #4

    gneill

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    Staff: Mentor

    The carts have different initial positions (only cart A starts at position 0). How would you modify your equation to account for this? Write the equation for each block.
     
  6. Oct 3, 2013 #5
    i set the both equations equal to each other, i also removed v1 from the equation as in both equations it is 0.

    so x=1/2at^2=x=1/2at^2.

    i also know that Cart B has to go negative for them to meet. But how would i account if for instance, cart b did not move, or vice versa? i also have to answer what would happen if both of them did not move. and my equation is supposed to answer all of these questions. thanks for the help so far.
     
  7. Oct 3, 2013 #6

    gneill

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    Staff: Mentor

    The problem is, you haven't accounted for the fact that cart B does not start at x=0. The equation for cart B requires a term for its initial position. Yes, the initial velocities are both zero, so those terms vanish.

    If either cart doesn't move then its acceleration will be zero. No problem! The math will take care of it. It just means that the cart that does move will eventually run into a cart that's standing still.
     
    Last edited: Oct 3, 2013
  8. Oct 3, 2013 #7
    Ok i got Xb= Xf-Xb for the cart B position term.
     
  9. Oct 3, 2013 #8

    gneill

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    Staff: Mentor

    Can you explain how you got that? Isn't Xb one of the unspecified givens (parameters) for the problem? And surely Xf is the final position that you're trying to locate.

    You need to work the initial position of cart B, which is a parameter of the problem, into the equation for the position of cart B. In other words, the version of the formula you've used to find the cart positions over time is not the most general version of that formula. The more general version has terms for both initial velocity AND initial position, so things don't all have to begin at the origin.
     
  10. Oct 6, 2013 #9
    Alright, how would i go about finding this formula? Sorry i'm not really keen on physics lol.
     
  11. Oct 6, 2013 #10

    gneill

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    Staff: Mentor

    Somewhere in your notes or textbook you'll have a list of kinematic equations. The one you've quoted already, (1/2)at2, is a trimmed-down version of the one you want -- you want the version that includes the initial position term.
     
  12. Oct 6, 2013 #11

    adjacent

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    Gold Member

    There are 4 kinematic equations
     
  13. Oct 6, 2013 #12
    ok would i use: vf^2= vi^2+2ax

    that's the only one i found with position in it. if so, how would i apply this?
     
  14. Oct 6, 2013 #13

    gneill

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    Staff: Mentor

    If your course text and notes are failing you, you could always try the Physics Forums' Introductory Physics Formulary. Look under the Basic Equations of 1-D Kinematics for Displacement & Time.
     
  15. Oct 6, 2013 #14

    Oh ok thanks a lot for that.

    so i would use: x=x0+v0t+(1/2)at2?

    What exactly does the 0 mean next to the x and v ? now that i have this how would i go about apply this?
     
  16. Oct 6, 2013 #15

    gneill

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    Staff: Mentor

    Yes. Note its similarity to the equation you first proposed. The difference is in the extra term representing the initial positional offset, ##x_0##.
    The "0" subscripts designate that they represent initial conditions. Sometimes an "i" for "initial" is used instead. Think of it as representing conditions that exist at time t = 0.

    As to how you apply it, that's what solving the problem is all about. What values would you put to each of the terms for the two carts involved?
     
  17. Oct 6, 2013 #16
    so for cart B: xf=xb+v0t+(1/2)aBt2
    and cart A: xf=0+v0t+(1/2)aAt2

    Do i need to set them equal to each other now?
     
  18. Oct 6, 2013 #17

    gneill

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    Staff: Mentor

    Both carts are starting from rest. What does that tell you about the ##v_0## parameters for each?

    Make sure that you adjust your equations to comply with the chosen coordinate system. You want cart B to be moving towards the left (towards the origin), so its acceleration term should be negative. Once that's settled, yes, equate them and carry on.

    P.S. You can create subscripts and superscripts using the x2 and x2 icons on the Advanced editing panel if you want to dress up your equations.
     
  19. Oct 6, 2013 #18
    xf=xb+v0t+(1/2)-aBt2=xf=0+v0t+(1/2)aAt2

    so like this?
     
  20. Oct 7, 2013 #19

    gneill

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    Staff: Mentor

    Sort of. Can you clean that up so that there's an equation (only one "=" sign), and do something about the minus sign in the middle of the cart B acceleration term; either place the "-aB" in parentheses or move the minus to the outside of the term. As it stands it looks like the minus sign is between two separate terms, which of course would be incorrect.
     
  21. Oct 7, 2013 #20
    xf=xb+v0t+(1/2)-(aB)t^2-0-v0t-(1/2)aAt2

    like this? still not sure how exactly to combine them together to have 1 = sign.
     
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