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Acura/Bentley: FINAL EXAM REVIEW

  1. May 6, 2012 #1
    1. Two cars, an Acura and a Bentley, are initially at rest a distance L from each other. At time t = 0 both cars starts accelerating toward each other, with constant acceleration Aa amd Ab respectively, until they crash into each other.

    a) At what time does the crash happen?

    b) Find the distance traveled by the Bentley before the crash.

    c) What is the relative speed of the Bentley, with respect to the Acura, just before the
    crash?


    Write your results in terms of L, Aa, and Ab. Remember to check the dimensions/units for
    each answer.




    2.
    EQUATIONS: (WE CANNOT USE MEMORIZED EQUATIONS, WE MUST DERIVE THEM GIVEN THE INFORMATION.

    Va = ∫Aa(dt) --> Aa(t) + Voa

    Xa = ∫Va(dt) --> 0.5Aa(t)^2 + Voa(t) + Xoa

    Vb = ∫Ab(dt) --> Ab(t) + Vob

    Xb = ∫Vb(dt) --> 0.5Ab(t)^2 + Vob(t) + Xob




    3. The attempt at a solution

    [a] The cars will crash when their positions are the same ---> Xa = Xb.

    0.5Aa(t)^2 + Voa(t) + Xoa = 0.5Ab(t)^2 + Vob(t) + Xob

    You can take out both Voa and Vob because the problem states that they are initially at rest. Let Xoa be at the origin, making it 0 and Xob = L because that is the distance between it and the Acura.


    ∴ 0.5Aa(t)^2 = 0.5Ab(t)^2 + L

    Aa(t)^2 - Ab(t)^2 = 2L ---> t^2 (Aa - Ab) = 2L

    t = [(2L/(Aa - Ab))]^0.5. Discard the negative solution for time.


    [b.] Find the distance traveled by the Bentley before the crash.

    Xb = 0.5Ab(t)^2 + Vob(t) + Xob

    Once again Vob(t) = 0,

    Xb = 0.5Ab([(2L/(Aa - Ab))]^0.5)^2 + L

    Xb = 0.5Ab(2L/(Aa-Ab)) + L

    Xb = [Ab(L)/(Aa-Ab)] + L

    Is this correct?

    Any help for this and part C will be greatly appreciated :)
     
  2. jcsd
  3. May 6, 2012 #2

    gneill

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    Staff: Mentor

    Check your assumptions about the signs of the various quantities. Suppose the accelerations of the vehicle's happened to be the same, what would be the time of the collision according to your derived formula?
     
  4. May 6, 2012 #3
    Because the Bentley will be accelerating to the left, it's acceleration should be labeled as -Ab.

    Which when plugged in should give:

    0.5Aa(t)^2 = -0.5Ab(t)^2 + L

    Aa(t)^2 + Ab(t)^2 = 2L ---> t^2 (Aa + Ab) = 2L

    t = [(2L/(Aa + Ab))]^0.5.
     
  5. May 6, 2012 #4

    gneill

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    Staff: Mentor

    That's better :smile: Always check to see if your results make sense by seeing if any 'pathological' behaviors can arise.

    Now you can rework your answer for part .
     
  6. May 6, 2012 #5
    Thank You :) I tend to loose marks due to these silly mistakes. Reworking Part B gives:

    XB = -0.5aBt2 + V0Bt + X0B

    V0Bt = 0

    ∴ XB = -0.5aBt2 + L

    XB = -0.5aB[(2L)/(aA+ aB]2 + L

    XB = -aB/(aA + aB] + L
     
  7. May 6, 2012 #6

    gneill

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    Staff: Mentor

    Okay, that's B's position, but how far did he travel? Distance traveled is his change in position.
     
  8. May 6, 2012 #7
    XB = -0.5aBt2 + V0Bt + X0B

    V0Bt = 0

    ∴ XB = -0.5aBt2 + L

    XB = -0.5aB[(2L)/(aA+ aB]2 + L

    XB = -aBL/(aA + aB] + L

    XB = -aBL/(aA + aB) + L(aA + aB)/(aA + aB)

    XB = aA(L)/(aA + aB)?

    I was looking over a document I found and in that document, the Bentley was placed at the origin instead of the acura, causing the final answer to be similar to the one above. Does it matter which car is placed at the origin?
     
  9. May 6, 2012 #8

    gneill

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    Staff: Mentor

    You've still only calculated the position of the Bentley. The distance covered should be given by the final position minus the initial position. Since the initial position is at distance L...
    No, it shouldn't matter. It will simply change the signs of some of the values.
     
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