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Solving for t in a distance equation.

  1. Dec 4, 2011 #1
    How do I get to this equation?
    [tex]t=\frac{v_f-v_i}{a}[/tex]


    Getting there from this equation is easy because there is only one t.
    [tex]v_i=v_f+at[/tex]


    But how do I get to it from this one?
    [tex]s = v_it+\frac{1}{2}at^2[/tex]


    Can someone help me with the steps. I am thrown off because there is multiple t's in the equations.
     
  2. jcsd
  3. Dec 4, 2011 #2

    gneill

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    Staff: Mentor

    Think: quadratic equation.
     
  4. Dec 4, 2011 #3
    [tex]ax^2+bx+c=0[/tex] ???

    In case you are wondering I am not in school I am studying on my own for fun. I have a journal and I am going through this physics book. It doesnt say anything about making sure I understand how to rearrange equations. I am doing this as an extra step the book doesnt even ask me to understand the rearrangement. I have been just taking them and rearranging on my own.

    This is what I thought is the first step but seems to over complicate it.

    [tex](v_i+\frac{1}{2}at)t[/tex]
    ??
     
  5. Dec 4, 2011 #4

    gneill

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    Yes, that's a quadratic. Quadratic equations generally have two 'roots' (solutions). That is, two values of x will make the equation balance. Sometimes both roots are the same (called a 'double root'), and sometimes they will be complex numbers. You have to choose the one that makes physical sense for the problem you're doing.

    The first step is to put your equation into the form above (in your case t takes the place of x). In this case you need to move the 's' over to join the other terms on the same side, leaving zero on the other. After than you can employ what is called the 'quadratic formula' to extract the roots. A google search on 'quadratic formula' will turn up useful info.
     
  6. Dec 4, 2011 #5
    Ok I am working on that now with the quadratic.

    [tex]t = \frac{-v_i ± \sqrt{v_i^2-4\frac{1}{2}as} }{2\frac{1}{2}a}[/tex]

    I am really not sure what I am supposed to do with the quadratic.
     
  7. Dec 4, 2011 #6

    gneill

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    Hang on. I just realized that you're not just trying to solve for t, you're trying to manipulate one equation into the form of the other.

    You're working with two different kinematic equations here. One is

    [itex] v_f = v_i + at [/itex]

    The other is

    [itex] s = v_i t + (1/2) a t^2 [/itex]

    Note that one includes a distance variable and the other doesn't. To get from one to the other requires a bit more than simple algebraic mucking about. It involves a bit of calculus.

    If you were to differentiate the equation [itex] s = v_i t + (1/2) a t^2[/itex] with respect to time, then you'd end up with the [itex] v_f = v_i + a t[/itex] equation.
     
  8. Dec 4, 2011 #7

    What does it mean to differentiate equation? What are the steps to differentiate?
     
  9. Dec 4, 2011 #8

    gneill

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    Differentiation is a calculus operation. It's not a simple algebraic manipulation in general. The topic of calculus is too involved for a simple answer here. You might want to do some research on the web.

    Fortunately there is another way to show the relationship between the formulas. You can go from the [itex]a = \frac{v_f - v_i}{t}[/itex] to the [itex]s = (1/2) a t^2[/itex] equation by geometrical means.

    If you have a graph of velocity versus time, then the distance covered by an object moving with that velocity is given by the area under the curve (between the curve and the time axis). This corresponds to your [itex]s = (1/2) a t^2[/itex] formula.

    For constant acceleration between velocities vi and vf, the curve is a straight line (the velocity increases at a constant rate). So the area under the curve is easy to calculate.

    attachment.php?attachmentid=41568&stc=1&d=1323062204.gif

    In the figure the line in red is the changing velocity. Note that its formula is given by:

    [itex] v(t) = v_i + a \; t[/itex]

    So "a" is the slope of the line. In order to determine the area under the curve it is convenient to divide the total into two portions, one rectangular and one triangular. The sum is then [itex] Area = v_i t + \frac{1}{2} (v_f - v_i) t [/itex]

    This should begin to look familiar!

    If we multiply the last term by t/t, which is just unity so it doesn't change the value, it then looks like:

    [itex] Area = v_i t + \frac{1}{2} \frac{(v_f - v_i)}{t} t^2 [/itex]

    and

    [itex] Area = v_i t + \frac{1}{2} a t^2 [/itex]


    And there you have your acceleration formula and distance formula all tied together.
     

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  10. Dec 4, 2011 #9

    NascentOxygen

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    If you know the values of everything on the right hand side, substitute them, and you have two values for t. Often you're lucky and only one of the values makes sense, so you can discount the other.

    Use (+) in the above equation to get one value of t, and use the (-) to calculate the other.
     
  11. Dec 4, 2011 #10
    Thank you gneill and NascentOxygen.

    gneill that post is exactly what I was hoping for. It helps me see what is going on... just perfect!
     
  12. Dec 5, 2011 #11
    how do I solve for t? What are the steps.

    [tex]s=v_it+\frac{1}{2}at^2[/tex]

    I am confused because of the binomial that has a t and a t squared on the other side.
     
  13. Dec 5, 2011 #12

    gneill

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    Staff: Mentor

    You can use the quadratic formula. Put the equation into the form:

    [itex] a t^2 + b t + c = 0 [/itex]

    then apply the quadratic formula to the coefficients a, b, c:

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

    You will get two "candidate" values for t. You will have to check to see if one or both are realistic solutions for the physical situation you're modelling.
     
  14. Dec 5, 2011 #13
    thank you gneil. I am trying to figure out how to algebraically rearrange the equation.

    I want to see the steps to get to the quadratic formula. So its not just a memorized formula.

    What would you do first? Then second...
     
  15. Dec 5, 2011 #14
    http://www.mathsisfun.com/algebra/quadratic-equation-derivation.html

    I found exactly what I was looking for here. Thanks guys.
     
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