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Distance and acceleration problem

  1. Aug 31, 2004 #1
    this picture shows the problem:
    http://s93755476.onlinehome.us/phys.jpg [Broken]

    i realize the discriminant must be positive for t_catch to have a real value. i just can't figure out how to write an expression with a and b. thanks in advance.
    Last edited by a moderator: May 1, 2017
  2. jcsd
  3. Aug 31, 2004 #2


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    Realizing that the discriminant must be greater than or equal to 0 is really the whole problem. The minimum speed, c, will give a discriminant of 0 (i.e., cmin2 - 2ab = 0). Just solve for cmin and plug back into your expression for tcatch from the quadratic formula.
  4. Aug 31, 2004 #3
    how do i solve for c_min?
  5. Aug 31, 2004 #4


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    cmin2 - 2ab = 0
  6. Aug 31, 2004 #5
    sorry im retarded, i didn't read your response correctly. thanks for your help!
  7. Aug 31, 2004 #6
    if you have a quadratic formula:
    [tex] ax^{2} + bx + c = 0 [/tex]
    the solution for its roots is:

    [tex] x = \frac {-b {+-} \sqrt{b^{2} - 4ac}}{2a} [/tex]

    now your equation is{
    [tex] \frac{1}{2}at^{2} - ct + b = 0 [/tex]

    so the solution to have a positive x walue would be easy to find, just sub in your constants in the equation above:

    [tex] x = \frac {-(-c) {+-} \sqrt{(-c)^{2} - 4((\frac{1}{2}a)(b)}}{2(\frac{1}{2}a)} [/tex]

    [tex] x = \frac {c {+-} \sqrt{c^{2} - 2ab}}{a} [/tex]

    so in order to have a positive x value, there must be a positive root,
    [tex] c^2 - 2ab > 0 [/tex]

    there is your answer. Do you understand now?
  8. Sep 1, 2004 #7


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    Hmm, I don't see the connection to the original question.
  9. Sep 1, 2004 #8
    what dont you see about it, the question is asking to express the minimum values of the mans speed in terms of a and b. As long as

    [tex] c^2 - 2ab > 0 [/tex]

    there is your answer. The min value would be:

    [tex] c^2 - 2ab = 0 [/tex]
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