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I need help in finding where two people will meet again

  1. Sep 16, 2015 #1
    1. The problem statement, all variables and given/known data
    At the instant the traffic light turns green, Stan Speedy starts from rest and accelerates at "as". At the same instant Kathy Kool passes Stan with constant velocity "vk".
    Given: [as, vk]

    How far beyond the starting point do they meet again?



    2. Relevant equations
    v = v0 + a*t
    x = x0 + v0t + 0.5 * a * t2
    v2 = v02 + 2a(x-x1)

    3. The attempt at a solution
    Stan:
    x0 = 0 <--- Since he started from rest.
    x = ?
    a = as <--- Given
    v0 = 0 <--- Stan started from rest
    v = v <----- I think velocity is equal to itself since as, Stan's velocity must be constant too.
    t = ?

    First I used v = v0 + a*t to find the time
    v = (0) + (as * t)
    t = v/as

    Now I substitute t into the position formula
    x = x0 + v0 * t + 0.5 * a * t2
    x = 0 + 0 + (0.5 * as * (v/as )2
    x = 0.5 * v2/as

    So now I have Stan's position.

    Going to list Kathy's variables:
    x0 = ? <--- from the wording of the prompt I assumed that Kathy has already been running before the green light.
    x = ? <--- somewhere in front of Stan
    v = vk <--- Given
    v0 = ?
    a = a <--- constant velocity means constant acceleration
    t = ?

    So my plan is to find Kathy's position and set it equal to Stan's position and search for the distance of where they meet again. However, I am pretty sure that my setup so far is incorrect. May someone please interject and set me straight?
     
  2. jcsd
  3. Sep 16, 2015 #2

    mfb

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    Things are always equal to themself. 3=3, 5=5, ...
    Stan is accelerating, his speed cannot be constant.

    The time of what? How (and why) do you want to find the time of anything of relevance without considering Kathy's motion here?

    It means more than that.
    Also, see above, a=a is a pointless statement. It is always true.

    That is good, but as you said, you have to fix the distances first.
     
  4. Sep 16, 2015 #3

    SteamKing

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    Constant velocity means zero acceleration.
     
  5. Sep 17, 2015 #4
    Thank you, you've cleared up my confusion on how acceleration cannot be constant. So now I am going to try to find Stan's and Kathy's position.

    Stan:
    x0 = 0 <--- Since he started from rest.
    x = ?
    a = as <--- Given
    v0 = 0 <--- Stan started from rest
    v = ?
    t = ?

    From here I will try to find Stan's position:

    x = x0 + v0*t + (1/2)at^2
    x = (0) + (0) + (1/2) at^2
    x = (1/2) at^2 <------ Stan's Position


    Going to list Kathy's variables:
    x0 = ? <--- from the wording of the prompt I assumed that Kathy has already been running before the green light.
    x = ? <--- somewhere in front of Stan
    v = vk <--- Given
    v0 = ?
    a = ? <--- constant velocity means constant acceleration
    t = ?

    Any idea where I start from here? Would it be correct if Kathy's x0 = 0m and v0 = 0m/s? I am confused on what to set her initial velocity and initial position to since in the prompt sounds like she was moving before the light even turned green.
     
  6. Sep 17, 2015 #5

    SteamKing

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    The last equation gives Stan's position x from the stop light as a function of t.

    If Kathy is traveling at a constant velocity of vk, how far has she traveled from the stop light in t seconds? Hint: d = r * t
     
  7. Sep 17, 2015 #6

    So Kathy's d = vk * t, meaning I need to find Kathy's time relative to Stan's and plug that into this, then I can get Kathy's position and finally set it equal to Stan's position.

    Is it correct to find "t" from Stan's position equation and plug it into Kathy's (d = vk * t)?
    Like: d = x = (1/2) at^2
    d = (1/2) at^2
    t = sqrt(2d/a)

    Kathy's position = vk * t
    = vk * sqrt(2d/a)
     
  8. Sep 17, 2015 #7

    mfb

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    You have two equations and two unknowns, the way to solve that is up to you. Your approach works, but setting the two expressions for the distance equal to each other is probably a bit easier.
     
  9. Sep 17, 2015 #8
    Going to list Kathy's variables:
    x0 = 0 <--- Kathy's initial position is zero because that's when she caught up to Stan
    x = ? <--- somewhere in front of Stan
    v = vk <--- Given
    v0 = ?
    a = 0 <--- constant velocity means acceleration = 0
    t = T

    So, I have made some changes to Kathy's key. I changed her x0 to 0 because that's where the origin (meaning where we'll began the calculations).
    I also changed her acceleration to 0 because it was giving that her velocity "vk" is a constant.

    Now, nothing has changed about Stan:
    Stan:
    x0 = 0 <--- Since he started from rest.
    x = ?
    a = as <--- Given
    v0 = 0 <--- Stan started from rest
    v = ?
    t = T

    Now, what I can see is that Kathy's time(t) and initial position (x0) are shared with Stan's.

    Where the work begins:

    Stan's Final Position:
    x = x0 + v0t + 1/2 * a*t^2
    x = 0 + 0 + (1/2) * (as) * T^2
    x = (1/2) * (as) * T^2

    Kathy's Final Position:
    x = x0 + v0t + 1/2 * a * t^2
    x = 0 + vk * T + 1/2 * (0) * t^2 <--- a = 0 because Kathy's velocity is constant
    x = vk * T

    Now I will set Kathy's and Stan's final position equal to each other.

    vk * T = (1/2) * as * T^2

    SOLVING FOR THE TIME:
    v = v0 + at
    v = 0 + as T
    T = v/"as" <---- Just a reminder that "as" is still Stan's acceleration.

    So far I am on the right path? How do I find how far beyond the starting point do they meet again? What should I do from here?
     
  10. Sep 18, 2015 #9

    mfb

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    Good. That equation is all you need now. T is the only unknown, you can find T.
    Where does that come from now and what does it mean?
     
  11. Sep 18, 2015 #10
    So to find T, I can either use the equation you quoted?

    v = v0 + at
    vk = 0 + at
    t = vk/a

    or can I

    vk * T = (1/2) * as * T^2

    vk = (1/2) * as * T
    T = 2vk/as

    and I am unsure when you asked "Where does that come from now and what does it mean?"

    Well...
    considering that acceleration is constant like in this problem,

    a = dv/dt

    adt = dv

    integrat:

    a | dt = |dv
    at +c = v

    and when time = 0
    v0 = vf(0) = a(0) + c

    v0 = c

    substitute c with v0

    v = v0 + at

    and it means that its an equation that is relative to time.

    Kathy (looking for time):
    vk = v0 + at
    vk = v0 + (0)t
    vk = v0

    so I am not sure how to solve for time with Kathy's information using v = v0 + at, unless it's valid that I make her v0 = 0, but I am unsure how or if its even valid to do so since the problem never mentioned her at rest.
     
    Last edited: Sep 18, 2015
  12. Sep 19, 2015 #11

    haruspex

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    That's all very confusing because you reuse symbols with different meanings, like v0, apparently as an initial velocity for both.
    Kathy has constant velocity, vk, so it is also her initial velocity. You can write down where Kathy is at time t (as you already did).
    Base on Stan's starting from rest and given constant acceleration, you can write down where he is at time t, as you already did. The final step is to say that these distances are equal at time T. You do not care what their velocities are at that time.
     
  13. Sep 19, 2015 #12

    mfb

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    The second approach works. The first one is randomly mixing equations together, that does not work.

    Good, you found the time. Where is Kathy at this time T? This will be the final answer.
     
  14. Sep 19, 2015 #13
    Thank you very much for your help! I need to do the same problem too, good thing that this was posted.
     
    Last edited: Sep 19, 2015
  15. Sep 19, 2015 #14
    Since it asks for "How far beyond the starting point do they meet again?", I should plug "T" into into Kathy's position "x = vkt" and that should give me the distance of how far beyond the starting point since I made both Stan's and Kathy's x0 = 0m?
     
  16. Sep 19, 2015 #15

    mfb

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    Right.
    You can also plug it into Stan's position, the result has to be the same by construction.
     
  17. Sep 19, 2015 #16
    Awesome, thank you! Do you mind walking me through a similar problem after I've posted what I've done on it in a little bit?
     
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