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Kinematics Car velocity Problem

  1. Aug 9, 2015 #1
    1. The problem statement, all variables and given/known data
    In the figure here, a red car and a green car move toward each other in adjacent lanes and parallel to an x axis. At time t = 0, the red car is at xr = 0 and the green car is at xg = 223 m. If the red car has a constant velocity of 20.0 km/h, the cars pass each other at x = 43.1 m. On the other hand, if the red car has a constant velocity of 40.0 km/h, they pass each other atx = 76.9 m. What are (a) the initial velocity and (b) the (constant) acceleration of the green car? Include the signs.



    2. Relevant equations

    q8GGHtZ.png

    3. The attempt at a solution
    eq 1 = 179.9=Vo(7.75)+1/2a(7.75)^2
    eq 2 = 146.1=Vo(6.92)+1/2a(6.92)^2

    insert 2 to 1

    Vo = 146.1-1/2a(6.92)/(6.92)

    179.9=(146.1-1/2a(6.92)/(6.92))(7.75)+1/2a(7.75)^2

    got a = 5.06 m/s^2
    and vo as 3.60

    is my answer right????
     
  2. jcsd
  3. Aug 9, 2015 #2

    mfb

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    Staff: Mentor

    You could explain where all those numbers in equation 1 and 2 come from, that makes it easier to understand the solution.
    Did you plug the result back into those equations to check?
     
  4. Aug 9, 2015 #3
    Vr = 20km/h = 5.56m/s x = 43.1
    v=d/t -> t=d/v = 43.1/5.56=7.75s
    xgreencar - x = 223 - 43.1 = 179.9

    179.9=vo(7.75)+1/2a(7.75)^2

    same procedure to eq. 2

    Sorry but I need a bit help here
    *is my answer right
    *if not, can you please help me?
     
  5. Aug 9, 2015 #4

    mfb

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    Staff: Mentor

    I suggested a way you can test your answer yourself.

    Yes it is right. Well, several units are missing.

    Edit: oh right, forgot about the signs.
     
    Last edited: Aug 9, 2015
  6. Aug 9, 2015 #5
    Check signs per original problem statement.
     
  7. Aug 9, 2015 #6
    That's one thing that bothers me I need signs how do I know what signs will I use? I just got a positive does it mean I'll use + sign?
     
  8. Aug 9, 2015 #7
    On the X axis,which way is the green car going?
     
  9. Aug 9, 2015 #8
    So 5.06 and 3.61 will be put a - sign on it? If that's what you meant
     
  10. Aug 9, 2015 #9
    Correct, if velocity and acceleration are in the (defined) negative direction, they will have negative values.
     
  11. Aug 9, 2015 #10
    Okay thank you!! Off to my next question in my next post see you there lol kidding Thankyouu
     
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