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Solving linear acceleration for time.

  1. Jan 18, 2012 #1
    I'm currently developing a program which solves the displacement of an object using the standard acceleration equation [d=i*t + 0.5*a*t^2], where i=initial velocity, t=time and a=acceleration.

    As the need has arisen to determine the time at which an object reaches a particular displacement (collision detection), I am currently using the equation [t = (sqrt(i^2 + 2ad) - i) / a] in order to solve for t.

    The problem is, I cannot for the life of me see how the second equation is derived from the first using standard algebraic manipulations. Whilst my program operates correctly using the aforementioned equations, it would be nice if someone could explain the manipulations involved in order to procure this derivation.

    Gemma.
     
  2. jcsd
  3. Jan 18, 2012 #2

    chiro

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    Science Advisor

    Hey GemmaT and welcome to the forums.

    For quadratic equations, the main step is to complete the square. Lets look at a standard quadratic equation:

    ax^2 + bx + c = 0. Want to solve for x.

    Now what we do is complete the square for the ax^2 + bx terms. Move c to the other side.

    ax^2 + bx = -c.
    x^2 + (b/a)x = -c/a (Divide both sides by a)

    This is where we complete the square. We know that (x + d)^2 = x^2 + 2dx + d^2 using standard expansion. What we want to do is match up the 2d with (b/a) which means 2d = b/a which means d = b/2a. Doing this we get:

    x^2 + (b/a)x = -c/a goes to
    x^2 + (b/a)x + (b/2a)^2 = -c/a + (b/2a)^2 (Add (b/2a)^2 to both sides)
    (x + (b/2a))^2 = (b/2a)^2 - (c/a) (Using result above)

    (x + (b/2a)) = +-SQRT( (b/2a)^2 - (c/a)) (Taking square roots and remembering positive and negative solutions)

    x = -b/2a +- SQRT((b/2a)^2 - (c/a)). Now factor out the 2a term and put it on the denominator and we get

    x = (-b +- SQRT(b^2 - 4ac))/2a which is the quadratic formula.

    Using the appropriate a,b,c for your problem should give you the right result.
     
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