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Homework Help: 1-D Kinematics Problem

  1. Oct 6, 2008 #1
    Beginner's Kinematics Problem

    1. The problem statement, all variables and given/known data
    An x-axis likes along an inclined plane, with the positive x-direction down the plane. Two objects A and B travel along the x-axis.

    At t=0, object A starts at x=0 with an initial velocity of 0. At t=1s, object B starts at x=10.00 m with a velocity of -5 m/s. Both objects have an acceleration of positive 2 m/s/s.

    At what time and position do the objects collide?

    At what time do the objects have the same speed?

    2. Relevant equations
    xf= x + vi(t) + 0.5at2

    vf= vi+a(t)


    3. The attempt at a solution
    a) Equate the two equations (because x final is the same) and solve for t.
    OBJECT A: xf= 0 + (0 m/s)(tf-0 s) + 0.5(2 m/s/s)(tf-0)2

    OBJECT B: xf= 10 m + (-5 m/s)(tf-1 s) + 0.5(2 m/s/s)(tf-1)2

    I got t= 2.29 seconds using a graphic calculator. Is that correct? Are the equations above correct?

    Part b of the question is what's REALLY confusing me. There is no time that I can find that the objects have the same speed.

    OBJECT A:
    vf= (0 m/s) + (2.00 m/s/s)(tf-0s)

    OBJECT B:

    vf= (-5 m/s/s) + (2.00 m/s/s)(tf-1s)

    The two lines, when graphed, never intersect. Also, making object A's equation equal to object B's equation doesn't work either. What am I doing wrong??
     
    Last edited: Oct 6, 2008
  2. jcsd
  3. Oct 6, 2008 #2

    LowlyPion

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    First of all observe that the equation for the first one released is simply:
    x = 1/2*a*t2 but it starts a second earlier so, ...
    x = 1/2*a*(t+1)2

    The second ball then can be described as:
    x = 10 -5*t +1/2*a*t2

    Since we are in the same time scale (having adjusted the first equation) they will meet when x = x.
    Substituting a=2 and solving for time then gives you the solution.
    (t+1)2 = 10 -5*t + t2

    2t + 1 = 10 - 5*t

    7*t = 9

    Part 2:

    v = a*(t+1)

    v = 5 - a* t

    2*a*t = 3

    4 * t = 3
     
    Last edited: Oct 6, 2008
  4. Oct 6, 2008 #3
    Haven't I already adjusted the time scale, though, by having the (tf-1 s) part of the second equation?

    EDIT: It seems that your solution is exactly one second behind my solution. How do I know which to use?
     
    Last edited: Oct 6, 2008
  5. Oct 6, 2008 #4

    LowlyPion

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    Yes, but I think you must not have been careful along the way.

    See the added edit, below for part 2

    Also your time scale puts the second ball leaving a second earlier I believe.
     
  6. Oct 6, 2008 #5
    So using (tf-1 s) is not wrong? Because I tried it again and it definitely gives the answer that I originally got.

    I think what's happening when you adjust for the time is that you're moving the origin so that x=0 is at x=1s. But the problem, I think is asking for the time that they collide where time was originally 0. So I think my original answer in that case might be correct?

    For the second part, I think I understand the first two lines, but not the last two:
    2*a*t = 4

    4 * t = 4

    Shouldn't I be able to equate the first two lines to find the time at which v is equal?
     
  7. Oct 6, 2008 #6

    LowlyPion

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    Oops. You are right. My way puts the frame in terms of t=0 when the second ball is released. So yes it's 1 second too fast. Sorry for any confusion.

    For the a*t terms its the same thing. But yes I set v = v then solved.
    Add 1 second to that as well.
     
  8. Oct 6, 2008 #7
    Ah okay, I think I get it now.

    Except the a*t terms thing still isn't working for me. Instead of having t+1 on the first equation, it should be just t now, right? So that means on the second equation I will have to have t-1.

    This should give:
    OBJECT A:
    vf= (0 m/s) + (2.00 m/s/s)(tf-0s) or vf=(2.00 m/s/s)(t)

    OBJECT B:

    vf= (-5 m/s) + (2.00 m/s/s)(tf-1s) or vf=(-5 m/s ) + (2 m/s/s)(t-1)

    When equated, there is no solution!

    (2t)= -7 + 2t ---> 2t-2t=-7 -----> 0=-7?
     
  9. Oct 6, 2008 #8

    LowlyPion

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    It asks for speed being equal, not velocity.

    Note I corrected a small math error in the first way I did it. 4*t = 3

    But you add 1 to adjust to the same as this way. Sorry again that my simple math errors may have confused. I was just scratching them out without being careful.
     
    Last edited: Oct 6, 2008
  10. Oct 6, 2008 #9
    Oh... I don't know how to calculate speed without giving it a positive or negative sign.

    I guess I still don't understand where you got 2*a*t = 4 from, either. Sorry, I'm taking physics for the first time this year and I'm not very good at it yet =)
     
  11. Oct 6, 2008 #10

    LowlyPion

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    On the one side you have v = a*(t) = 2 t

    On the other your speed starts out at 5 and slows to deceleration so = 5 - 2*(t-1)

    2t = 5 - 2t +2

    4 t = 7
     
  12. Oct 6, 2008 #11
    Oh, so when the velocity is -5.00 m/s, that means the speed is 5.00 m/s.

    Since the object is slowing down, however, the acceleration could be considered -2 m/s/s.

    I get it now!

    Thanks for all your help!!
     
  13. Oct 6, 2008 #12

    LowlyPion

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    Cheers then.
     
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