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0 = t(7.8 x 10² m/s - 4.9 m/s² t)

  1. Nov 5, 2007 #1
    [SOLVED] 0 = t(7.8 x 10² m/s - 4.9 m/s² t)

    1. The problem statement, all variables and given/known data


    0 = ( 7.8 x 10² m/s ) x t + ½ x (-9.5 m/s²) x t²

    I have time, represented by t in the equation twice....how would I figure out what time is? This is probably simple math, but being out of school for 7 years now, someone please refresh my memory.


    2. Relevant equations

    0 = t(7.8 x 10² m/s - 4.9 m/s² t)
    0 = t

    or 0 = (7.8 x 10² m/s - 4.9 m/s² t)

    4.9 m/s² t = 7.8 x 10² m/s

    t = (7.8x10²m/s) / (4.9 m/s²)
    = 1.6 x 10² s


    3. The attempt at a solution
     
  2. jcsd
  3. Nov 5, 2007 #2
    I didn't understand you, you already have the two values for time. These represent the two times at which your condition is met.
     
  4. Nov 5, 2007 #3
    you have your two times.... they're written right there
     
    Last edited: Nov 5, 2007
  5. Nov 5, 2007 #4
    opps...sorry guys....forgot to mention, the answer shown is not my answer, it's the answer in the back of the book.

    I don't understand how they got it. I only know how to isolate 't' (time), if there's one variable. i.e. 1 = t + 2.....isolated would be 1 - 2 = t.

    In the original equation, there's 't' twice......how would I isolate that?
     
  6. Nov 5, 2007 #5
    If we leave the equation in a symbolic form and factor out a t we get:

    0 = t(v+0.5gt)

    When does the RHS = 0?

    ax^2+bx=0 you could consider the equation like that if it helps...
     
  7. Nov 5, 2007 #6

    CompuChip

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    Let's drop all the units for a while, then your equation is of the type
    [tex]0 = a t^2 + b t[/tex]
    where a, b are just numbers and you want to solve for t.

    You can use the "no-thinking-involved" way and just plug it into the abc-formula, which says that the solutions for
    [tex]0 = a t^2 + b t + c [/tex]
    in general are
    [tex] t = \frac{ - b \pm \sqrt{b^2 - 4 a c} }{2 a}[/tex]
    (if you choose the plus or the minus you get two solutions).

    But in this case there is an easier way, and that's noticing that you can bring a t outside the brackets:
    [tex]0 = (a t + b)t,[/tex]
    if you work out the brackets you'll see that this is the same as the original equation.
    Now you have a product of two things, namely t and a t + b. When is a product of two numbers zero?
     
  8. Nov 5, 2007 #7
    now I'm more confused....what's v? and what's gt?
     
  9. Nov 5, 2007 #8
    Your equation 0 = ( 7.8 x 10² m/s ) x t + ½ x (-9.5 m/s²) x t²

    was derived from [tex]x_f =x_0+v_0t+\frac{1}{2}gt^2[/tex]

    v and g are just constants (7.8 x 10² m/s for v and -9.5 m/s² for g)
     
    Last edited: Nov 5, 2007
  10. Nov 5, 2007 #9
    where does c come from?

    and how does 0= at2 + bt + c

    work out to be


    [tex] t = \frac{ - b \pm \sqrt{b^2 - 4 a c} }{2 a}[/tex]

    ???


    i've never learned this math before, is there a name for this? I tried googling, but don't know what this is called.
     
  11. Nov 5, 2007 #10
    It's called the quadratic formula -- basically it's used to solve quadratic equations (equations that have a variable to the second power t^2)

    What's that highest level of math you've taken?
     
  12. Nov 5, 2007 #11
    grade 12 ..... but that's 7 years ago.

    been dicking around ever since, forgot everything.

    i'm taking Uni prep Physics grade 12 right now.

    cool....quadratic formula....going try google, see if there's a tutorial.
     
  13. Nov 5, 2007 #12
    Ah ok... Well there's an easier way then using that formula:
    0=(7.8 x 10² m/s) x t + ½ x (-9.5 m/s²) x t²

    Would you agree that we can factor out a "t" and get:

    0=t((7.8 x 10² m/s) + ½ x (-9.5 m/s²) x t)

    Would you also agree the anything multiplied by 0 will equal 0?

    So if t=0 we get:

    0=0((7.8 x 10² m/s) + ½ x (-9.5 m/s²) x 0)
    0=0(7.8 x 10² m/s + 0)
    0=0

    So we know that t=0 is a solution. But that's not the only solution.

    Would you agree that 0=t((7.8 x 10² m/s) + ½ x (-9.5 m/s²) x t) when

    (7.8 x 10² m/s) + ½ x (-9.5 m/s²) x t = 0

    since that would give us 0 = t x 0 which is 0
     
  14. Nov 5, 2007 #13

    CompuChip

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    Ah, since you were asking a physics-based question, I assumed a high-school level of mathematics. So I guess you have some reading up to do :smile: (as a side note, if you want to really learn algebra thoroughly, solving quadratic equations may not be the easiest way to start. )
     
  15. Nov 5, 2007 #14

    my heads spinning on the physics alone. I'm gonna try to figure out quadratic equations before I attempt other methods.
     
  16. Nov 5, 2007 #15

    yes, lots of reading......sigh...!!

    before this lesson, I ran into something I didn't know.....trigonometry. Had to download ebooks to figure it all out. I got most the basics down now.

    now quadratic equations.....sigh!!
     
  17. Nov 5, 2007 #16
    I'd like to comment on how wicked this forum is. Thnkx guys/.....I really appreciate all the help. I hope this forum stays as active as it is now.

    cheers.....got lots of reading to do.
     
  18. Nov 5, 2007 #17
    got a little backround, but a little stuck.


    0 = ( 7.8 x 10² m/s ) x t + ½ x (-9.8 m/s²) x t²


    using [​IMG]

    should be ?

    a= -4.9 m/s² t²
    b= 7800 m/s
    c= ???

    what would c be for this formula? would I just leave it out?
     
  19. Nov 5, 2007 #18
    a = -4.9 m/s^2
    b = 780 m/s
    c = 0 -- since we don't have a "c term"
     
  20. Nov 5, 2007 #19
    wicked!!!!!!!!


    it works!!!!!!!!! problem solved...!!!!!!!!!!!!!!!!!!!!!!
     
  21. Nov 6, 2007 #20

    CompuChip

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    The advantage of that formula is that it always works for a quadratic equation. All you have to do is bring it to the form [itex]ax^2 + b x + c = 0[/itex] by adding/subtracting terms on both sides of the equation, identify the a, b and c and plug them in.

    The disadvantage is, as you might have noticed by now, that some formula's are easier to solve without it. For example, you can solve
    [tex]a t^2 + b t = 0[/tex]
    by noticing that you can bring a t outside the brackets (as in the problem you originally posted):
    [tex](a t + b) t = 0 \Rightarrow t = 0 \text{ or } a t + b = 0 \text{ so } t = - b / a[/tex]
    or
    [tex]t^2 - c = 0[/tex]
    by just bringing the c to the other side and taking square roots.

    But if you don't see such a method, the quadratic formula always works.
     
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