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Average Accleration.

  1. Sep 1, 2003 #1
    This question was part of a 2 part question. I figured out the first part of it. I can't get the 2nd half of it. The formula for a lightweight object, such as a styrofoam ball is v=v_t(1-e^kt) where V_t=5.70 m/s is the terminal speed, k=1.72s^-1 is the drag coefficient of the styrofoam ball, t is the clock reading, in seconds, and e is 2.7183.., the base of the natural logarithms. What is the magnitude of the acclelration of the ball at t=0.5s? What is the average acceleration of the ball during the time intercal from t=0 to t=0.5 s?

    I figured out the magnitude of the acceleration of the ball at t=0.5 is 4.15 m/s^2. For the second part of the question I derived the formula up top making it k*e^(-k*t)*v_t. I plugged in the values and used 5.0 for t and did it again using 0 for t. I then subtracted the 2 values I got for t and ended up with 5.654. I can't figure out what I did wrong.
    Last edited: Sep 1, 2003
  2. jcsd
  3. Sep 2, 2003 #2
    You can do that only if you have an equation like that of the straight line: y=a*x+b. Otherwise you have to use:
    yavg=[1/(x2-x1)] * ∫x1x2 dx y(x)
  4. Sep 2, 2003 #3


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    Staff Emeritus
    Science Advisor

    Okay, since you got the first part correct, you recognize that you need to differentiate the velocity function to get the acceleration function (and you did that correctly).

    Sonty is correct that, in general, in order to find the average of a continuous function, you have to integrate between the two limits and then divide by the difference: the average acceleration between t= 0 and t= 0.5 is (int(t=0 to 0.5) a(t)dt)/(0.5- 0).

    HOWEVER, since (by the fundamental theorem of calculus) the integral of the acceleration function is the velocity function, this is exactly the same as (v(0.5)- v(0))/0.5= -4.82 m/s^2.
    Your mistake was that you forgot to divide by the difference in times. (Your "solution", subtracting the two velocities, gives m/s. You need to divide by time to get m/s^2.)
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