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When will these cars pass eachother - kinematics

  1. Feb 19, 2006 #1
    Car A with u=0/s is on the northbound land of a highway and car B is travelling in th southbound lane at constant u=26.67m/s.At t=0 , A starts and accelerates at a constant rate a(A) . while at t=5 , B decelerates with constant magnitude of a(A)/6 . knowing that when the two carspass each other at x=90m , and v(A) = v(B),determine acceleration of car A.

    i have tried it many times but stil can't ind the answer.pls help.
    the answer is 3.59m/s^2
    thanx
     
  2. jcsd
  3. Feb 19, 2006 #2
    oh...the length between the two cars is d in which we have to find in the second part
     
  4. Feb 19, 2006 #3
    You just need to write the equations of velocity and movement of each car. Then you'll find out the answer
     
  5. Feb 19, 2006 #4
    for v(A)=180a(A)--------1
    s(A)=1/2 at^2--------2

    v(B)=26.67 - 0.5a(B)t--------3
    d-90=26.67t - 0.5a(B)t^2-----4

    is it rite??
     
  6. Feb 20, 2006 #5
    pls help.........
     
  7. Feb 21, 2006 #6
    pls help........
     
  8. Feb 21, 2006 #7
    Teng! I'm not a moderator but all you have to do is ask ONCE! :surprised

    Tricky! I like it.

    In the following, I have defined + to be Northward.
    For car A: Car A starts at position [tex] x_{0A} [/tex] at a speed of [tex] v_{0A} = 0m/s[/tex] with an acceleration of a(A).
    [tex] x_A=x_{0A}+v_{0A}t+(1/2)a_At^2 [/tex]; [tex]v_A=v_{0A}+a_At [/tex]
    [tex] x_A=x_{0A}+(1/2)a(A)t^2 [/tex]; [tex]v_A=a(A)t [/tex]

    For car B: Car B starts at position [tex] x_{0B} [/tex] at a speed of [tex] v_{0B} = -26.67 m/s[/tex] with an acceleration of -(1/6)a(A).
    [tex] x_B=x_{0B}+v_{0B}t+(1/2)a_Bt^2 [/tex]; [tex]v_B=v_{0B}+a_Bt [/tex]
    [tex] x_B=x_{0B}-26.67t-(1/12)a(A)t^2 [/tex]; [tex]v_A=-26.67-(1/6)a(A)t [/tex]

    Now, at x=90 m, both cars pass each other and have the same speed. We don't have a specific equation to find t or a(A), but we have two equations we can write. First, at time t both cars have the same speed. Thus at (90 m, t):
    [tex] a(A)t=-26.67-(1/6)a(A)t [/tex]

    Second, both cars are at the same point at this time, so:
    [tex] x_{0A}+(1.2)a(A)t^2=x_{0B}-26.67t-(1/12)a(A)t^2 =90 m [/tex]
    (This last is actually THREE equations. Both distances are equal to 90 m as well as each other.)

    You've got four equations in four unknowns, so you should be able to solve the system.

    -Dan
     
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