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Back with more questions

  1. Dec 15, 2004 #1
    Thanx for everyone who has helped so far.
    (This is not homework, I'm doing problems to study for my final that is in less than 3 hours).
    This next question i'm having a hard time with is...

    A 2150-kg truck is traveling along a straight,level road at a constant speed of 55.0km/h (15.8m/s) when the driver removes his foot from the accelerator. After 21.0 seconds, the truck's speed is 33.0 km/h (9.2m/s). What is the magnitude of the average net force acting on the truck during the 21.0 second interval?

    The answer is 626 N but I have no idea on how to get to it.
     
  2. jcsd
  3. Dec 15, 2004 #2
    This is one dimensional movement along a straight line...

    [tex]v_f = v_i - a * t[/tex]. where i and f denote initian and final...

    just try it and multiply with m to find the force...


    regards
    marlon
     
  4. Dec 15, 2004 #3
    Marlon, that doesn't seem right, I keep getting 675 and the answer says its 626 (multiple choice question, closest answer is 626, and teacher said answer is 626)
     
  5. Dec 15, 2004 #4
    yes, i see your point though i am quite certain of this...

    regards
    marlon...

    perhaps someone else can help...
     
  6. Dec 15, 2004 #5
    F = dp/dt = m(v_2 - v_1)/ t = 2150kg(9.2m/s - 15.8m/s)/21s ~ 675

    675 has to be right. Your book/teacher is wrong :P
     
  7. May 19, 2010 #6
    you have your conversions wrong.. 55 km/h is 15.3 m/s not 15.8 m/s. The rest is correct.
     
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