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Homework Help: Ave/Instantateous Velocity

  1. Sep 16, 2005 #1
    Hello Everyone,

    Here is a question that is so stupid I am embarrassed to ask it, I know it should be easy, but for some reason I can't get it :grumpy:

    1) An object starting from rest has a constant acceleration of a=8m/s^2. Find the average velocity from t=0 to t=10s using the definition of ave velocity. Show that it equals instantaneous velocity at the midtime t=5s. (Hint use the forumulae for constant a)



    Ok, so I know that average velocity is change in position over change in time, and that average acceleration is change in velocity over change in time. I also know by to find the instantaneous velocity I must use change in position over change in time of the point immediately before and immediately after the midpoint (5). Now for some reason I can't seem to get anywhere with this, I hate the beginning of the school year my mind is always so useless :blushing:

    Thanks for any suggestions you can give me to get me on the right track.
     
  2. jcsd
  3. Sep 16, 2005 #2

    robphy

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    This is really an algebra problem.
    First write what you have to show...
    [tex]\displaystyle\frac{x_{10}-x_{0}}{t_{10}-t_{0}} \stackrel{?}{=} v_{5}[/tex]
    Use the hint: "the formulae for constant acceleration". (My hint: use one of these formulas on the left for [tex]x_{10}[/tex], and use the other on the right.)
     
  4. Sep 17, 2005 #3
    Ok, I'm afraid I'm still a little bit lost. Would it be possible to say that if acceleration is 8m/s^s, take the integral of that to get velocity? No wait, that wouldn't work because that would give me instantaneous velocity right? I think I'm still a bit confused.
     
  5. Sep 18, 2005 #4

    robphy

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    It's fine. Since the acceleration is constant, the instantaneous velocity has a simple form... which you will use on the right-hand side. By doing another integral, you'll get an expression for the instantaneous position that you'll use on the left-hand side. Suggestion don't use the specific value of 8m/s^2 now... stick to the algebra and keep it as "a".
     
  6. Sep 18, 2005 #5
    So do you think it would work it I went ave v=instantaneous v which equals (change in pos/change in time)= (v^2/2)?

    I'm also not sure how to find the average velocity when you are given time but no position.
     
  7. Sep 18, 2005 #6

    Fermat

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    If an object starts from rest and moves with a constant acceln of 8 m/s², how far does it travel in 10s?
     
  8. Sep 18, 2005 #7
    Ok, the average velocity is just 8m.s^2 x 10s = 80m/s


    So to show that the ave v is equal to inst v;

    We now know that the average velocity is 80m/s. If this is true we can find how far it travels in 5s. In 5 seconds we find it travels 400m. So we can now put the numbers into both sides of the equation:

    (80m/s^2)(10s)=(400m)/(5s)
    80m/s=80m/s

    Is this correct?
     
  9. Sep 18, 2005 #8

    Fermat

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    Nope.
    Sorry, but that is the final velocity.

    One of the eqns of motion is,

    vf = vi + at

    where vf is the final velocity
    vi is the initial velocity
    a is the accln
    t is the time of travel

    Since we're starting from rest, then vi = 0. The accln = 8 m/s² and t = 10s, so
    vf = 0 + 8*10
    vf = 80 m/s
    =========

    Getting back to the problem though, you have to find the distance travelled in 10s. You should use the formula,

    [tex]x_f = x_i + v_i*t + \frac{1}{2}at^2[/tex]

    [tex]\mbox{Here,}\ x_i = 0\ \mbox{and}\ v_i = 0,\ \mbox{so the formula reduces to,}[/tex]

    [tex]x_f = \frac{1}{2}at^2[/tex]

    Putting in a = 8 and t = 10, you will get

    [tex]x_f = 400m[/tex]
    =========

    [tex]\mbox{But}\ x_f\ \mbox{is the distance travelled at }t = 10[/tex]
    [tex]\mbox{i.e. }x_f = x_{10}[/tex]
    [tex]x_0 = 0[/tex]
    [tex]t_{10} = 10[/tex]
    [tex]t_0 = 0[/tex]
    [tex]v_{av} = \frac{(x_{10} - x_0)}{(t_{10} - t_0)}[/tex]
    [tex]v_{av} = \frac{(400 - 0)}{(10 - 0)}[/tex]
    [tex]v_{av} = 40 m/s[/tex]

    [tex]\mbox{Now you have to use one of the eqns of motion to show that the instantaneous velocity at }t = 5,\ v_5,\mbox{ has the same value as }v_{av}[/tex]
     
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