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Help with Kinematics and Acceleration

  1. Oct 10, 2008 #1
    1. The problem statement, all variables and given/known data

    A car initially traveling at 15 m/s accelerates at a constant rate of 4.5 m/s^2 over a distance of 45 m. How long does it take the car to cover this distance?

    Vi: 15 m/s
    Vf: ___ m/s
    A: 4.5 m/s^2
    Delta D: 45 M
    Delta T: ___ s

    2. Relevant equations

    I'm guessing I would have to use this Kinematics equation:
    Delta D (Displacement) = Vi * Delta T (Time) + 1/2 * A * Delta T^2

    3. The attempt at a solution

    I tried attempting this problem but had trouble rearranging the equation to solve for the missing variable T.

    Could someone give me a basic rundown on HOW TO rearrange equations as I never really learned and or understood this concept.

    Lastly, what are some good classical/intro physics text books I can buy/download as my school assigned us Glencoe's Physics : Principles and Problems.
  2. jcsd
  3. Oct 10, 2008 #2
    It's going to be quadratic, so you depending on how convinient your values are (not very convinient in this equation) you can solve it with normal factorising methods, or use the quadratic formula.

    However if this is too difficult/confusing. You can always use:

    [tex] V_f^2 = V_i^2 + 2as [/tex]

    and then sub final velocity into:

    [tex] V_f = V_i + at [/tex]
  4. Oct 10, 2008 #3
    what is s?
  5. Oct 11, 2008 #4
  6. Oct 11, 2008 #5
    It shouldn't be that hard to use the quadratic formula to solve the equation you posted.

    Show us how you substituted the given values into that equation
  7. Oct 11, 2008 #6
    truth is i dont know how to use the quadratic forumla to solve this problem, i've never done something like this >_>
  8. Oct 11, 2008 #7
    Well, you've already got:

    45 m = 15 m/s * t + 1/2 * 4.5 m/s^2 * t^2

    The quadratic formula applies to an equation of the form:
    [tex]ax^2 + bx + c = 0[/tex]

    In that equation, the unknown is x, in your equation the unknown is t. Can you match up the constants a, b, and c with the constants in your equation?
  9. Oct 11, 2008 #8
    Oh yeah sorry I'm a bit unorthodox, I use s as displacement.

    Just to confirm that you know the formula. [tex] \frac{-b\frac{+}{} \sqrt{b^2 - 4ac}}{2a} [/tex]
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