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Homework Help: Physics problem I cant get

  1. Sep 4, 2004 #1
    1.5 = (0.25 m/s)t +(0.48m/s2 )t 2

    How do I find what t is???

    If you know can you please show me how to do this!!!

    This question is relating to a Physics problem that my lab group is stump on

    My Aim is Atarikid4000, Thanks. :smile:
  2. jcsd
  3. Sep 4, 2004 #2


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    You have 0.48 t2+ 0.25 t- 1.5= 0.

    In general, at2+ bt+ c= 0 has solutions
    [tex]t= \frac{-b+\sqrt{b^2-4ac}}{2a}[/tex] and
    [tex]t= \frac{-b-\sqrt{b^2-4ac}}{2a}[/tex].

    In your problem, a= 0.48, b= 0.25, c= -1.5.
    Last edited by a moderator: Sep 4, 2004
  4. Sep 4, 2004 #3
    I recommend writing a program in your graphing calculator that automatically will solves using the quadratic formula. Saves time.

    My $.02

    Paden Roder
  5. Sep 4, 2004 #4
    Thanks for the help!!!! :smile:
  6. Sep 4, 2004 #5


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    One more thing:

    You probably already "got most of the physics" when you arrived at that equation (assuming it's correct)! [That's great!] Given your equation, "solving for t" is a math problem... not a physics problem. The only physics is deciding which of the two roots to accept for your answer.

    I tell my students to try to distinguish
    "problems with physics" (e.g., what is going on?, how do I set up the problem?, am I using the appropriate physical laws and definitions?, how do I interpret my answer?)
    "problems with math" (e.g., how do I solve for x?, what is the volume of a cylinder?, should I use sin or cos?, what is the x-component of this vector?).
    This helps the student (and me) zero-in on where the real problems lie.

    (It's a peeve of mine that "physics" is often mistakenly blamed for problems in math.)
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