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Dragster average acceleration

  1. Sep 13, 2004 #1
    A piston-engine dragster set a world record by starting from rest and hitting a top speed of 244 mi/h over a measured track of 440 yd. Compute its average acceleration in m/s^2.

    I assumed the acceleration to be constant in order to find t using this formula : v_av = 1/2(v_i + v_f).
    After that used this one : a= deltaV/deltaT.
    My answer is 15 m/s^2 but the answer given in the text book is 18 m/s^2.
    What's wrong? My assumption?

  2. jcsd
  3. Sep 13, 2004 #2
    The speed is given to you in miles per hour. You need to convert this to meters per hour, then to meters per second. I'll assume you've done all this.

    Your error comes in using the formula,
    v_av = 1/2(v_i + v_f).

    [tex]v(average) = \frac{1}{2}at^2[/tex]

    This only works if you know the average velocity. In this question, we only know the maximum velocity, not the average one.

    You know the initial velocity (0), the maximum (or final) velocity and the distance. So use one of the equations of motions to find the acceleration. You don't even need to worry about the time if you use this one:

    [tex]v^2 = u^2 + 2as[/tex]

    where a=acceleration, v=final (maximum) velocity, u=initial velocity (0) and s=displacement in meters, not yards. Now you can work out your acceleration!

    Hope this helped!

    AMW Bonfire

    P.S. If it still doesn't work, make sure you've converted all units to standard units, ie m/s and meters, etc.
    Last edited by a moderator: Sep 13, 2004
  4. Sep 15, 2004 #3
    I followed what you said, but got the same answer :cry:
    My answer was, as is now, 15 m/s^2 but the answer in the solutions manual is 18 m/s^2.
    This the way that problem is solved in the book :

    a_av= deltaV/deltaT = (244 * 0.4470 m/s) / ( 6.2 s) = 18 m/s^2 .

    Where the heck the 6.2 s comes from ?
  5. Sep 15, 2004 #4
    Hm... where did the "delta time" come into it?

    I'm honestly not sure. Can anyone else help with this?

    AMW Bonfire
  6. Sep 15, 2004 #5


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    "delta time" means "change in time" and here is the time from start to finish of the race. That's not given but you can calculate it. Assuming a constant acceleration for the entire race, since the initial speed was 0 and final speed 244 mph, the "average" speed was 122 mph. 440 yards is 1/4 mile so the time required to go 0.25 miles at an average speed of 122 mph is 0.25/122= 0.00204918 hours which is (multiply by 3600) 7.377 seconds. Hmm, that is NOT "6.2 seconds".

    There are 1609 meters per mile, approximately. That means one "mile per hour" is 1609 "meters per hour" and 244 mph is 392596 meters per hour or (dividing by 3600 seconds in an hour) 109 meters per second. To go from 0 to 109 meters per second in 7.4 seconds requires an acceleration of approximately 15 m/s2, just as you got initially. There may be an error in the book- check with your teacher on this one.
  7. Sep 15, 2004 #6
    I'm reading physics on my own, there is no teacher to check with.
    PF is the only place to ask my questions.
    Thank you very much amwbonfire and HallsofIvy
  8. Sep 16, 2004 #7
    Sorry, I misworded my question. I understood how to calculate the change in t, but I didn't understand why it is a requirement to work out the answer. Of course, I see now there are multiple ways of solving the problem. Thanks! :smile:

    That explains why you ask a lot of questions. :wink: The people on PF are always friendly enough to help though. Keep asking away! After all, he who asks is a fool for five minutes, he who does not is a fool forever.

    AMW Bonfire
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