Thread: If you were the teacher... View Single Post
HW Helper
P: 2,952
 Quote by Galileo Well, as unfair as it may sound afterwards, the obtained answer is not the most important thing. The reasoning behind it is. The journey is more valuable than the destination (or something like that). The average velocity is by definition the displacement divided by the time interval: $$\bar v = \frac{\Delta s}{\Delta t}=\frac{s(b)-s(a)}{b-a}$$ This is (by definition) general. It always holds. Constant acceleration or not. By constrast, if the acceleration is constant, there is another way. he velocity graph is a straight line, to find the average you can simply take the average of the end velocities. But this is not general. It doesn't holds for, say s(t)=t^3 from t in [0,1]. Therefore, your answer is incomplete without specification that the acceleration is constant, which you could show by calculating s''(t)=-32 (=constant). So in all fairness, you got a little bit less than the full 2 points. 1.5 is not bad, right? Consider you learned from it.
I'd like to add the definition for the accepted meaning of "average value of a function".

Say you have a function f(x) defined and integrable over an interval [a,b]. The average value of that function over that interval is defined as :

$$f_{avg} = \frac{1}{b-a} \int_{a}^b f(x)dx$$

That applies for *any* function. If you take the same equation and apply it to a problem of motion to calculate average velocity over the interval [t1, t2], you get :

$$v_{avg} = \frac{1}{t_2-t_1} \int_{t_2}^{t_1} v(t)dt$$

Keeping in mind that v(t) = ds/dt that becomes :

$$v_{avg} = \frac{1}{t_2-t_1} \int_{t_2}^{t_1} \frac{ds}{dt}dt$$

$$v_{avg} = \frac{1}{t_2-t_1} [s(t_2) - s(t_1)]$$

or, in other words, the total displacement divided by the total time. This method made no assumptions about the acceleration, and therefore can be applied in all circumstances.

You'll be seeing more of average value of a function in physics, notably in a.c. circuit theory.