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Approaching vehicles with acceleration

  1. May 15, 2017 #1
    1. The problem statement, all variables and given/known data
    Two vehicles, A and B, are approaching each other. Vehicle A has a constant velocity of ##v_A##. Vehicle B has a constant acceleration ##a## with an initial velocity of ##v_B##. The starting distance at ##t=0## between them is ##D##.

    Is there a way to solve at what time ##t## they'd meet?

    2. Relevant equations

    ##D = v_At + v_Bt + \frac{1}{2} \cdot a \cdot t^2##

    3. The attempt at a solution
    I deduced the equation above but I can't find a way to solve for t. Simplifying gives:

    ##\frac{D}{t} = v_A + v_B + \frac{1}{2} \cdot a \cdot t##

    Is there a way to solve for t without plotting these two equations?
     
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  3. May 15, 2017 #2

    Ray Vickson

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    It is just a quadratic equation in ##t##, so is easily solved using the standard quadratic-solution formulas.
     
  4. May 15, 2017 #3

    PeroK

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    Your equation looks very like a quadratic in ##t## to me!
     
  5. May 15, 2017 #4
    Ah, for some reason I thought this wasn't possible because there are two parameters with either a ##t## or a ##\frac{1}{t}## in it. But after looking at it I noticed it could be written as:

    ##\frac{1}{2} \cdot at^2 + (v_A + v_B)t - D = 0##

    And then use the abc-formula in which ##a=\frac{1}{2}\cdot a## and ##b=(v_A + v_B)## and ##c = D##. Is this correct?
     
  6. May 15, 2017 #5

    PeroK

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    You already had the quadratic in section 2 of your original post.

    I guess you mean ##c = -D##?

    You may like to think what difference it makes if vehicle A is at rest and vehicle B is initially moving at the combined speed ##v_a + v_b##.
     
  7. May 15, 2017 #6
    Apologies, I did mean ##c = -D##. I take it you mean that I could just put the equation in section 2 in parentheses? My initial target was to use the abc-formula to only give the positive solution of ##t##.
     
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