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Physics/Calculation of Average Force

Question
Mr. Kovalcin,

I teach AP physics at The Woodlands High School in Texas.  I made up a problem in which I give the students a simple equation for velocity as a function of time: v(t) = 4 + 3t^2 and the mass of the object exhibiting this motion (10 kg).  I then ask the students to determine multiple values from this function, including the work done (delta KE) from t=0 to t=4 and the impulse (delta p) from t=0 to t=4.

I run into a problem, though, when I tried to calculate the average force over that time interval.  Using integration, I could find the displacement of the object for that time interval.  Thus, I thought I could find average force using two methods:
1.  Divide work done by total displacement
2.  Divide impulse by total time
and that these two calculations would yield the same results.  They do not, however.

Which method, if either, is accurate?  Where is my flaw in the incorrect method?

Thank you.

The flaw in your analysis is that the resulting force is not linearly proportional to time. As a result the average does not lie midway between the end points.
As an example I would like to use a somewhat simpler velocity function to reduce the mathematical complexity - this simplification will not change the outcome.
I am going to assume that the velocity function is : v(t)=3*t^2
Taking the anti derivative the displacement function will be:
D(t)=3*t^3/3=t^3  (assuming the displacement is 0 at t=0)
Taking the derivative of the velocity function the acceleration function,will be:
a(t)=6*t
And the force function will be:
F(t)=m*a=10*6*t=60*t
Evaluating each of the these functions at t=4s:
a=6*4=24m/s^2
v=3*4^2=48m/s
d=t^3=4^3=64m
F=60*4=240
Calculating the work done is given by:
W=Integral[F dot dx]
In this case dx can be determined from the displacement function from above:
x=t^3  which becomes  dx=3*t^2dt
So the integral for the work done becomes:
W(4) = Integral [F dot 3*t^2 dt] = Integral [60*t*3*t^2 dt] = Integral [180*t^3 dt] = 180/4*t^4 (4) = 45*4^4 = 11520J
Average force would then be given by:
Fave = W/d = 11520J/64m = 180N
Since the average force developed from the initial and final forces (at 0s and 4s) will be:
Fave = (0+240)/2 = 120N
Why are these answers different? Because the force as a function of time is non-linear!
F(t) = 180*t^3
And the average of the function, since it is a curve, does not lie on the midpoint of the curve.
Questioner's Rating
 Rating(1-10) Knowledgeability = 10 Clarity of Response = 10 Politeness = 10 Comment Very quick and helpful. I really appreciate the time given to assist me.

Physics