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Homework Help: For Dummies S.A.D. Question

  1. Nov 10, 2011 #1
    1. The problem statement, all variables and given/known data
    A drag racer’s acceleration is 26.6 meters/second2, and at the end of the race, its final speed is 146.3 meters per second. What is the total distance the drag racer traveled?

    2. Relevant equations
    Vf^2=1/2(26.6)(146.3)^2= 409 meters

    3. The attempt at a solution

    The attempt at the solution is my problem.. if any of you are familiar with physics for dummies workbook and read this question on p.34 it says that the equation above = 409 meters.. My questions is how?!

    When i enter 1/2(26.6)(146.3)^2 into my calculator i get 284669.077.. Which is a long way off from the answer the book gives.

    I am not currently a student in a math program I just find math fascinating but this has been bugging me for weeks. I was hoping someone could show me what i am missing
  2. jcsd
  3. Nov 10, 2011 #2


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    Welcome to physicsforums!

    Could you write down an expression for the car's position, s, as a function of time, t? That would be the best place to begin this problem.

    - Warren
  4. Nov 10, 2011 #3
    Thank You, Warren

    Actually, s is what I am trying to solve i guess it would be. the formula according to the book s=1/2a Vf^2 = Vf^2 = 1/2(26.6)(146.3)^2


    s= ?
    a= 26.6
    Vf^2 = 146.3
  5. Nov 10, 2011 #4


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    I think there's a parsing issue with the equation. It should read:
    [tex] s = \frac{1}{2 a} v_f^2 [/tex]

    It derives from the "standard" kinematic equation,
    [tex] v_f^2 - v_i^2 = 2 a s [/tex]
    when vi = 0.

    The numerics are dodgy, too. The result should be closer to 402m, not 409m.
  6. Nov 10, 2011 #5
    Yes thank you.. thats how it looks.. but between 402m and 409m i cant get anywhere near those numbers

    I am just trying to understand how it equals that
  7. Nov 10, 2011 #6


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    Staff: Mentor

    You must be suffering calculation order-of-operation issues. What do get for v2? (show us). Next divide that by 2. What have you got so far?. Next divide that by a. What's the result?
  8. Nov 10, 2011 #7


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    Yes, I think there's an error in the book, or you've copied the equation down incorrectly.

    Take a close look at the equation you posted, [itex]s = \frac{1}{2} a v_f^2[/itex]. The units don't work out -- meters/second^2 times meters^2/seconds^2 certainly does not come out to be meters.

    I always start these kinds of problems in one of two places. If I'm given a value for time, then I use this equation, which is straightforward:

    [tex]s(t) = s_0 + v_0 t + \frac{1}{2} a t^2[/tex]

    If I'm not given a value for time, then I start with Torricelli's equation, which can be derived from the above. It's independent of time, so makes things a little easier:

    [tex]v_f^2 = v_i^2 + 2 a s[/tex]

    In this case, I would start with Torricelli's.

    - Warren
    Last edited: Nov 10, 2011
  9. Nov 10, 2011 #8
    I believe v^2 is already accounted for as 146.3

    so if i went 146.3/ 26.6 i get 5.5

    The whole question and answer is already written out in the book.
    What i am not understanding is how the book got the answer it did..

    If i enter it in a calculator i enter 1 Divided by 2 multiplied by 26.6 multiplied by 146.3^2 and i end up with a huge number..

    I am debating making a video of what i am seeing and doing since i am not familiar with how to make the formulas on here like gneill
  10. Nov 10, 2011 #9


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    dvsj67, if I rearrange Torricelli's equation to solve for position, s, I can get this expression:

    [tex]s = \frac{v_f^2}{2 a}[/tex]

    If I plug in 146.3 m/s for [itex]v_f[/itex], and 26.6 m/s^2 for a, I get s = 402.3 meters.

    As gneill suggested, to enter this in a calculator, you must be careful with the order of operations. You could do the following:

    146.3 * 146.3 = 21403.69
    answer / 2 = 10701.845
    answer / 26.6 = 402.325

    You may wish to use a more sophisticated calculator that supports parentheses, or even https://www.google.com/search?gcx=c...,cf.osb&fp=7ccfaba8b72cb7e1&biw=1358&bih=739"!

    - Warren
    Last edited by a moderator: Apr 26, 2017
  11. Nov 10, 2011 #10
    Thank You, Warren

    That actually clarified this for me, enough that after a week of writing and rewriting this out I am confident to try a similar question
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