1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Subject of formula

  1. Jun 1, 2013 #1

    adjacent

    User Avatar
    Gold Member

    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

    User Avatar
    Gold Member
    2016 Award

    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

    User Avatar
    Science Advisor
    Homework Helper

    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

    User Avatar
    Gold Member

    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

    User Avatar
    Science Advisor
    Homework Helper

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

    adjacent

    User Avatar
    Gold Member

    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

    User Avatar
    Science Advisor
    Homework Helper

    i don't understand :confused:

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

    adjacent

    User Avatar
    Gold Member

    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

    User Avatar
    Science Advisor
    Homework Helper

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

    adjacent

    User Avatar
    Gold Member

    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

    Staff: Mentor

    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

    User Avatar
    Science Advisor

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

    Mentallic

    User Avatar
    Homework Helper

    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.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Subject of formula
Loading...