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Drag racing problem

  1. Jan 15, 2008 #1
    I have a problem in my D.E. class that is driving me nuts.

    Two drivers, A and B are in a race. Beginning from a standing start they both proceed at a constant acceleration. Driver A covers the last 1/4 of the distance in 3 seconds. Driver B covers the last 1/3 of the distance in 4 seconds. Who wins and by how much time?

    I am assuming the distance is 1320ft or 1/4 mile.

    I know that driver B wins by .594 seconds but I don't know how that number was reached. Can anyone help?
     
  2. jcsd
  3. Jan 15, 2008 #2
    Use s = ut+1/2.a.t.t
    separately for each car

    s = 1/4mile for both, and u = 0 (initial speed) for both
     
  4. Jan 16, 2008 #3
    Let the total distance of the race to be x.

    Then,

    Formula used: [tex] s = ut + \frac{1}{2}at^2[/tex]

    Consider the overall race:

    Driver A: [tex] x = \frac{1}{2}a_at_a^2...(1)[/tex] since the initial speed, u is zero.

    Driver B: [tex] x = \frac{1}{2}a_bt_b^2...(2)[/tex] since the initial speed, u is zero.

    Next,

    Find the velocity of A when reaches 3/4 of the distance (just before entering the last 1/4 of the distance)

    Find the velocity of B when reaches 2/3 of the distance (just before entering the last 1/3 of the distance.

    Formula used: [tex] v^2 = u^2 + 2as [/tex]

    Consider from the beginning of the race until the above mentioned point.

    Driver A: [tex] v_a^2 = 2a_a(\frac{3}{4}x)[/tex] since the initial speed, u is zero.
    Simplified :
    [tex] v_a^2 = \frac{3a_ax}{2}...(3)[/tex]

    Driver B: [tex] v_b^2 = 2a_b(\frac{2}{3}x)[/tex] since the initial speed, u is zero.
    Simplified :
    [tex] v_b^2 = \frac{4a_bx}{3}... (4)[/tex]

    Next,
    consider the last part of the race.

    Formula used: [tex] s = ut + \frac{1}{2}at^2[/tex]

    Driver A : [tex] \frac{1}{4}x = \sqrt{\frac{3a_ax}{2}}(3) + \frac{1}{2}a_a(9)[/tex]

    Solve for [tex]a_a \ in \ term \ of \ x:
    a_a = 0.0039887x; \ \ \ \ 0.77379x...(5)[/tex]

    Driver B : [tex] \frac{1}{3}x = \sqrt{\frac{4a_4bax}{3}}(4) + \frac{1}{2}a_b(16)[/tex]

    Solve for [tex]a_b \ in \ term \ of \ x:
    a_b = 0.41246x; \ \ \ \ \ 0.0042092x...(6)[/tex]

    Next,

    put (5) into (1),

    [tex] t_a = 22.3915\ s\ for \ a_a = 0.0039887x; 1.6077 s \ for \ a_a = 0.77379x \ which \ is \ not \ possible \ since \ the \ last \ part \ has\ already \ taken \ 3 s.[/tex]

    put (6) into (2),

    [tex] t_b = 21.7979\ s\ for \ a_b = 0.0042092x; 2.2020 s \ for \ a_b = 0.41246x \ which \ is \ not \ possible \ since \ the \ last \ part \ has\ already \ taken \ 4 s.[/tex]

    Since [tex]t_b[/tex] is shorter, he is the winner. He is faster by 22.3915 - 21.7979 = 0.594 s
     
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