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Homework Help: Need help for a mechanic problem

  1. Sep 26, 2004 #1

    mad

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    Hi guys.. this is my first post. Sorry if I use terms that you don't use since I translate it from french..

    I need help doing this problem. I tried it a few times but I can't find the answer. I tried to do the V(x) (speed or velocity I think) and find it from there but still can't manage to have it.. so if someone could help me.. this is the problem:

    A car at rest accelerates uniformly for 200m. She then continues at constant speed for 160m and decelerates 50meters before stopping. The entire movement lasts 33seconds.

    Question: how much time was she at constant speed ?

    Sorry for my english..
    thanks for the help
     
  2. jcsd
  3. Sep 26, 2004 #2
    I believe you need another variable. You have the distance, you want to find the time but what is the speed???

    Acceleration(m/s²) = Change in Velocity(m/s) / Time Taken for Change (s)

    or a = (v-u)/t

    This is one equation you are likely to need.

    The other is:

    Speed(m/s) = Distance Travelled(m)/Time Taken(s)

    or s = d/t

    There we go for equations.

    The Bob (2004 ©)
     
    Last edited: Sep 26, 2004
  4. Sep 26, 2004 #3

    mad

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    Thanks, but without the time taken, I cant use those 2 formulas.. ill try and post later.
    thanks
     
  5. Sep 26, 2004 #4

    arildno

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    This can be dealt with in the following manner:
    1. Let the speed at the constant level be V
    2. In the period of uniform accelerations, the average velocity (distance traveled/time spent) is V/2
    I'll prove that if you like..
    3.Now let t1 be the time spent in acceleration, t2 the time spent at constant speed, and t3 the time spent in deceleration.
    Hence:
    V/2*t1=200
    V*t2=160
    V/2*t3=50
    Or, by division:
    t1=2*(200/160)t2=5/2t2
    t3=2*(50/160)t2=5/8t2

    Or, since t1+t2+t3=33, we get:
    33/8*t2=33, that is t2=8
     
  6. Sep 26, 2004 #5

    mad

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    Wow, thanks a lot! How are you able to do that so easily? Do you have a trick or something ;)

    I just have a question: why is the average velocity v/2 ? (.2)
    Velocity = distance/time.
    You know it travelled 200m but dont know the time it spent.. could you explain me how you found v/2 ?

    thank you
     
  7. Sep 26, 2004 #6

    arildno

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    I'm old, but not too senile yet :wink:

    I'll argue for the average value as follows, when we start initially at rest:
    1. Having some constant acceleration "a", we know that the velocity v(t) at time "t" can be written as:
    v=at
    2. We also know that the distance traveled r(t) is: r= at^{2}/2

    3.Rewriting, we have:
    r=(at/2)t=(v/2)t
    4. Hence, the average velocity r/t equals v/2; that is, half the final velocity.
     
  8. Sep 26, 2004 #7
    This is what I was trying to get at but I couldn't. I feel quite useless now. Oh well..... that is life. The young give way to the older way of thinking.

    The Bob (2004 ©)
     
  9. Sep 26, 2004 #8

    arildno

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    But only for a short time..very soon I'll descend past you into senility and decrepitude.
     
  10. Sep 26, 2004 #9
    Not ncessarily. Just live well. :biggrin:

    The Bob (2004 ©)
     
  11. Sep 26, 2004 #10

    arildno

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    Thx!
    You seem to be a promising young man with possibly a bright future in front of you (aargh, that one should be in "Lame Jokes")
     
  12. Sep 27, 2004 #11
    :cry:

    :rofl: :rofl: :rofl: :rofl:

    The Bob (2004 ©)
     
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