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Mathematical problem is classical question

  1. Feb 14, 2009 #1
    From the formula 2as = v2 - u2

    I made v the subject...and it becomes -

    (2as+u2)1/2 = v

    Problem is real world value of 2as is negative...as a result it makes a complex answer...but actually its not; instead the value of v should also come negative.

    Here a, u are negative...and a force applies on the body in the same direction as u, which is negative, as a result a negative a cause of that very force.

    So the v should also be real and negative...but its coming as complex :confused:

    Say...can I take the a and u to be positive initially and add the negative sign to the result...sounds ok to me.
  2. jcsd
  3. Feb 14, 2009 #2
    [tex]v=\pm \sqrt{2as+u^2}[/tex]
  4. Feb 14, 2009 #3


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    How about the sign of s?
  5. Feb 15, 2009 #4
    Yeah its too negative.

    But that way the value that comes by is positive, but it should be negative.
    Last edited: Feb 15, 2009
  6. Feb 15, 2009 #5
    See Phrak's response
  7. Feb 15, 2009 #6


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    Show this explicitly in a specific problem and we will show you where you forgot another sign somewhere. Both Phrak and atyy have given you sufficient hints.

  8. Feb 15, 2009 #7

    Doc Al

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    When 2as is negative that means the body is slowing down. But a negative value of 2as cannot have a magnitude greater than u²--once the body slows to zero it must reverse direction.
  9. Feb 16, 2009 #8
    It is accelerating, but all the coordinates are negative, I mean...the u, a and s are all negative.
  10. Feb 16, 2009 #9
    Ok then...both negative and positive results will be valid?...I mean...that's true when a value is squared.
  11. Feb 16, 2009 #10

    Doc Al

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    So then what is the issue? 2as is positive, thus v² = u² + 2as is positive. No complex answers required.

    (Realize that this equation only gives you the magnitude of v; the sign is up to you.)
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