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Homework Help: Instantaneous v = average v

  1. Feb 21, 2010 #1
    instantaneous v = average v (solved)

    1. The problem statement, all variables and given/known data

    Indicate moments of time when the average velocity is equal to instantaneous velocity.

    Given values

    t (s) 0 | 2 | 5 | 7 | 8
    x (m) 0 | 4 | 16 | 12 | 11


    t (s) 0 | 2 | 5 | 7 | 8
    v (m/s) 0 | 4 | 4 | 0 | -2

    2. Relevant equations

    X = At^2 + Bt + C
    V = At + B

    V(int) = lim delta x / delta t as t approaches zero


    3. The attempt at a solution

    There are two problems before this.

    [a] Find average velocity of the object during first 5 second. 16/5 m/s
    Find the average acceleration of the object during the same period. 4-0/5 = 4/5

    I also find out the average velocity for each noticeable time intervals.

    0 - A : 4-0/2-0 = 2 m/s
    A - B = 16-4 / 5-2 = 4m/s
    B - C = 12-16/7-5 = -2m/s
    C- D = 11-12/8-7 = -1m/s

    One way to find the instantaneous velocity is to take the slope of the tangent line on position vs time graph.
    Another way is to find out the function of each interval and take ds/dt

    So, in general, X(t) = At^2 + Bt + C

    If we take V = At + B
    let t = 0, and V is also 0,

    V = At, and when t = 1, V = 2, 2 = A(1), A = 2, so for velocity of 0 - A time interval, I see the function is
    V = 2t

    This is true for 0 - A, B - C and C - D interval (i also think C - D is just part of B - C)

    For A - B, it is constant velocity, so it is just a constant, V = 4

    Now, coming back to solve [c] find instantaneous, I see X = At^2 + Bt + C

    I use similar technique, and try to find a function for each time interval. If I do, I take dx / dt, and plug in the time, to find the instanteous velocity at any time.


    I also know that constant velocity at any time = its instantaneous, but what about the rest.

    So far, is my approach correct? I feel like I am doing too much here.
     
    Last edited: Feb 22, 2010
  2. jcsd
  3. Feb 22, 2010 #2
    pushed, any thoughts is appreciated.
     
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