Q:

Prove that at some point the acceleration has a magnitude of at least 4 units

A particle moves in a straight line starting from rest and finishing at rest, and covers unit distance in unit time. Prove that at some point its acceleration has a magnitude of at least 4 units. It is assumed that $v(t)=\frac{dx}{dt}$ and $a(t)=\frac{dv}{dt}$ are continuous functions of time.
The proof given is the following:If we plot $v$ against $t$, the area under the curve must be the same (1 square unit) as that of an isosceles triangle having the same base and an altitude of 2.

The slopes of the sides of the triangle are $\pm 4$. Part of the $v-t$ curve must fall outside the triangle or coincide with its sides. Thus the slope, $a$, of the curve is $\geq 4$ at some point.
I was looking for a pure analytic proof which doesn't rely on the graphical representation and intuition.

A:

Let $v(t)$, $a(t)$ be the velocity and acceleration at time $t\in [0,1]$. We express distance as the integral of velocity and integrate by parts, giving
$$
\begin{eqnarray*}
1&=&\int_0^1 v(t)\,dt\\
&=&-\int_0^1 a(t)\left(t-\frac{1}{2}\right)\,dt.
\end{eqnarray*}
$$
Suppose $|a(t)|<4$ for all $t$. Then
$$
\begin{eqnarray*}
-\int_0^1 a(t)\left(t-\frac{1}{2}\right)\,dt&\leq&\int_0^1 |a(t)|\left|\left(t-\frac{1}{2}\right)\right|\,dt\\
&<&\int_0^{\frac{1}{2}}4\left(\frac{1}{2}-t\right)\,dt+\int_{\frac{1}{2}}^{1}4\left(t-\frac{1}{2}\right)\\
&=&\frac{1}{2}+\frac{1}{2}=1.
\end{eqnarray*}
$$

