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Does this involve calculus? im confused help please

  1. Sep 14, 2003 #1
    Hint: x_1 and x_2 never have the same value. The x-coordinates of two objects moving along the x-axis are given below as a function of time t. x_1 = (4m/s)t x_2 = -(25m) + (8m/s)t - (2m/s^2)t^2 Calculate the magnitude of the distance of closest approach of the two objects.
  2. jcsd
  3. Sep 14, 2003 #2
    Hi pringless,
    no, you do not need calculus to solve this.
    Let's call x(t) = x1(t) - x2(t).
    This is just quadratic in t (the graph is a parabola).
    All you got to do is find the lowest (or highest) point of the parabola.
    Let's call that point (t0, x0), then we can write
    x(t) = a(t - t0)2 + x0.
    You can find a, t0, x0 by matching the coefficients on both sides. OK?
  4. Sep 14, 2003 #3
    im sorry...i dont really understand what u mean
  5. Sep 14, 2003 #4


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    Graph the two equations. That will show you position of each particle as a function of time.

    If you then subtract one from the other, you'll have the difference between the two. If you graph that, you'll see the difference as a function of time. You'll see that it will go down and then go back up. The closest approach is where the difference is the smallest.
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