So I was working on a very basic physics problem that had something to do with finding the height of a triangle(the velocity vs time graph) enter image description here
A body starts from rest with an acceleration of 2m/s^2 till it attains the maximum velocity ,the decelerates to rest with 3m/s^2, If the total time taken was 10 secs, What was the maximum velocity? I was able to solve it pretty fast and deduce that the height is 12.
Then I had another idea , What If Without changing the height and area/the total displacement I could make the triangle symmetrical? and divide t into two equal parts?
Then I could multiply the slope with the "run" and get the rise aka Vmax. The mistake I made was thinking the average of the slope after the transformation would still be the same constant. From (2,-3) to (2.5 and -2.5).And I had good reason to assume that, because one inch taken from one line gets added to the other and the same goes for the area. Why not the slope?
Just imagine a 45/45/90 triangle. The first (upward) leg has a slope of 1, while the other leg has a slope of -1.
Shift the vertex to the left so the upward leg has a slope of 2. But the downward leg certainly doesn't have a slope of 0.
But we can certainly see that as the absolute value of one slope gets bigger the other one gets smaller. Is there a way to know at what rate one slope changes in relation to the other?
You have three points of interest in your diagram: $(0,0), (t_2,0), (t_1,v_{\text{max}})$. The area $A$ of the triangle is given by $A = \frac{t_2 v_{\text{max}}}{2}$.
Let's rewrite $A$ in terms of the two slopes you describe. This will give the desired relation for constant $A$. The first slope is $a_1 = \frac{v_{\text{max}}}{t_1}$. The second slope is $a_2 = \frac{v_{\text{max}}}{t_1 - t_2}$.
Thinking geometrically, \begin{align} A &= \frac{t_1 v_{\text{max}}}{2} + \frac{(t_2 - t_1)v_{\text{max}}}{2} \\&= \frac{v_{\text{max}}^2}{2a_1} - \frac{v_{\text{max}}^2}{2a_2} \\&= \frac{v_{\text{max}}^2}{2}\left( \frac{1}{a_1} - \frac{1}{a_2} \right) \end{align} Thus, the desired relation is $$ A = \frac{v_{\text{max}}^2}{2}\left( \frac{1}{a_1} - \frac{1}{a_2} \right) $$ Note that the height of the triangle is $v_{\text{max}}$ and the displacement is given by the area $A$.
Why is always a slippery question, but the flaw in your reasoning is essentially that area varies directly with time whereas acceleration varies inversely with it (in this problem). This is why the relation above assumes the form it does, with displacement depending on the harmonic mean of the accelerations, not their average or something like that.
A Little More
I notice you make specific mention of a relationship between the rates at which $a_1$ and $a_2$ change. You can, of course, pass a derivative to figure that out. Holding $A$ and $v_\text{max}$ constant and differentiating implicitly, we find: $$ 0 = \frac{v_\text{max}^2}{2}\left(-\frac{1}{a_1^2} + \frac{1}{a_2^2} \frac{da_2}{da_1} \right) $$ which yields the (perhaps satisfying) relationship: $$ \frac{da_2}{da_1} = \left(\frac{a_2}{a_1}\right)^2 $$ Note that $A$ and $v_\text{max}$ drop out of this expression. We could have anticipated this since $\frac{1}{a_1} - \frac{1}{a_2}$ is a constant under the variations of the problem we are considering and does not depend on $A$ or $v_\text{max}$.