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Kinematics - Motion along a straight line

  1. Sep 29, 2006 #1
    I’m having difficulty with an assignment for my Physics class.

    Here’s the question:

    “Cars A and B are racing each other along the same straight road in the following manner: Car A has a head start and is a distance [tex]D_{A}[/tex] beyond the starting line at t = 0. The starting line is at x = 0. Car A travels at a constant speed [tex]v_{A}[/tex]. Car B starts at the starting line but has a better engine than Car A, and thus Car B travels at a constant speed [tex]v_{B}[/tex], which is greater than [tex]v_{A}[/tex].

    How long after Car B started the race will Car B catch up with Car A?
    Express the time in terms of given quantities.”

    My embarrassingly unsuccessful attempt:
    My thinking was that, when Car B catches Car A, [tex]x_{A} = x_{B}[/tex].

    I substituted the given quantities for Car A and Car B separately into the equation,
    [tex]x = x_{o} + v_{ox}(t-t_{o}) + \frac{1} {2}*a_{x}(t-t_{o})^2[/tex]

    x = position as a function of time
    xo = initial position
    vox = initial velocity
    t = a certain time
    to = initial time
    ax = acceleration

    For Car A, I got (simplified):
    [tex]x = D_{A} + v_{A}(t)[/tex]

    For Car B, I got (also simplified):
    [tex]x = v_{B}(t)[/tex]

    Equating them, [tex]D_{A} + v_{A}(t) = v_{B}(t)[/tex], and solving for t, I got:
    [tex]\frac{v_{B}t - D_{A}} {v_{A}} = t[/tex]

    Which is incorrect.

    I know I shouldn’t have “t” on the left side, but other than that I’m completely lost and would really appreciate it if anyone could point me in the right direction.
    Last edited: Sep 29, 2006
  2. jcsd
  3. Sep 29, 2006 #2
    Having issues fixing the fraction in my answer, so I'll try again here.

    Here is what I got:

    [tex]\frac{v_{B}t - D_{A}} {v_{A}} = t[/tex]
  4. Sep 30, 2006 #3


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    It may be easier to consider thier relative velocity ([itex]v_{r} = v_{B}-v_{A}[/itex]).
  5. Sep 30, 2006 #4
    Thanks a bunch, got the answer!
  6. Oct 1, 2006 #5
    I have the same question with the cars. I've made car A equal to car B and was able to get this equation:

    Da + va(t) = vb(t) and then i subtracted va(t) and then took out the common factor of t like so:

    Da= t(vb-va)

    then found that t= Da/(vb-va)

    I'm not sure if this is the right equation, i'm pretty sure it's not so if anyone can help me that would be great!
  7. Oct 2, 2006 #6


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    What makes you think your answer is incorrect? If your question is the same as the one originally asked then your solution is correct.
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