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Subject of formula

  1. Jun 1, 2013 #1

    adjacent

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    NOT a home work
    A person moves with constant speed of 10m/s(Initially as well as finally). His acceleration is therefore zero.Distance moved is 200meter.Find time taken.I tried to use the equation below but had difficulty making t the subject
    S=ut+1/2at2
    a=Acceleration
    t=time
    u=speed
    S=Distance moved
    If I use the values given I could easily make t the subject
    That is 200=10t+1/2*0*t2
    Giving 200=10t (t=200/10) that is t=s/u in this case

    I cant make t the subject using letters ONLY.There should be a way
     
  2. jcsd
  3. Jun 1, 2013 #2

    phinds

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    If a man travels for 2 hours at 60 mph, how far does he go? What equation do you need to solve that question?
     
  4. Jun 1, 2013 #3

    tiny-tim

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    hi adjacent! :smile:
    if a ≠ 0, this is a quadratic equation, which you can solve with t = -b ± √(b2 - ) etc

    if a = 0, it's simply s = ut, so t = s/u :wink:
     
  5. Jun 1, 2013 #4

    adjacent

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    S=ut+1/2at2
    That means t cannot be made the subject of the formula by methods I use with (S=1/2(u+v)t)Etc?
    What if S=ut+1/2at2 is given and asked to solve for t?
     
    Last edited: Jun 1, 2013
  6. Jun 1, 2013 #5

    tiny-tim

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    yes, solve it as a quadratic equation!!
     
  7. Jun 1, 2013 #6

    adjacent

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    For example no values for a,u,tetc were given.I solve it as a quadratic,If a turn out to be 0 then It would be wrong.Right?Meaning the equation is wrong for non accelerating objects(when for t)but when solved for S it is right even for non accelerating objects.How is it?I need proof
     
    Last edited: Jun 1, 2013
  8. Jun 1, 2013 #7

    tiny-tim

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    i don't understand :confused:

    show us your solution for the quadratic​
     
  9. Jun 1, 2013 #8

    adjacent

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    For example,
    a=0
    s=200m
    u=10m/s
    t=20s
    I can solve the equation even when a=0,(for s)
    S=ut+1/2at2
    =10*20+1/2*0*202 =200m:smile:
    But for t,
    0.5at2+ut-s=0
    t=(-b+√(b2-4ac)/2a
    t=(-10+√(102-4(0)(-200))/(2*0) = MATH ERROR:confused:
    In general If acceleration=0 - cannot solve for t
    If acceleration≠0 - can solve for t
    If acceleration is either ≠or= -can solve for S
    Why?
     
  10. Jun 1, 2013 #9

    tiny-tim

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    because the quadratic formula (-b ± √etc) does not work for a = 0
     
  11. Jun 1, 2013 #10

    adjacent

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    But if it can be solved for S,There should be some method to solve for t too.(Except t=s/u which is obtained once you know the value of a)
    Just Algebra(Only letters)?
     
  12. Jun 2, 2013 #11

    Mark44

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    tiny-tim answered your question in post #3. There are two cases: a = 0 and a ≠ 0.

    If a = 0, then t = s/v.
    If a ≠ 0, then you can solve for t by using the quadratic formula.
     
  13. Jun 2, 2013 #12

    pwsnafu

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    A quadratic equation means that a≠0.
    ##0x^2 + 2x + 4## is not a quadratic.
     
  14. Jun 2, 2013 #13

    Mentallic

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    I see what you're saying, and the answer to it is that the quadratic expression given by t = ... isn't quite complete because it assumes [itex]a\neq 0[/itex]. What we need is a piecewise function to describe it

    If [itex]s = ut + 1/2at^2[/itex]

    Then

    [tex]t =
    \begin{cases}
    \frac{-u\pm \sqrt{u^2+2as}}{a}, & a\neq 0 \\
    \frac{s}{u}, & a=0
    \end{cases}[/tex]

    You need to also keep in mind that if the discriminant is less than zero, which is [itex]a\neq 0[/itex] but [itex]u^2+2as<0[/itex] then you won't have a real value for t. Physically, this means that the object won't ever cross the line at displacement s because it would be accelerating away from that direction.
     
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