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

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 $$ \begin{eqnarray*} -\int_0^1 a(t)\left(t-\frac{1}{2}\right)\,dt&\leq&\int_0^1 |a(t)|\left|\left(t-\frac{1}{2}\right)\right|\,dt\\ &<&\int_0^{\frac{1}{2}}4\left(\frac{1}{2}-t\right)\,dt+\int_{\frac{1}{2}}^{1}4\left(t-\frac{1}{2}\right)\\ &=&\frac{1}{2}+\frac{1}{2}=1. \end{eqnarray*} $$

Here's a more general result:

Let $f:[a, b] \to \mathbb{R}$ is differentiable, not constant and $f(a) =f(b) =0.$ Show that there exists $c \in (a, b)$ such that $|f'(c)| > \dfrac{4}{(b-a)^2}\displaystyle \int_{a}^{b}f(x) dx$

Proof: The result is immediate if $f'$ is not bounded on $[a, b].$ If not then $|f'|([a, b])$ is non-empty and bounded above and hence has supremum $M >0.$ Then for all $x \in [a, b], $ $|f'(x)| \leq M.$ If $x \in [a, \frac{a+b}{2}]$ then by the mean value theorem applied to $f$ on $[a, x]$ there exists $c_x \in (a, x)$ such that $\dfrac{f(x) -f(a)}{x-a} =f'(c_x)$ and since $f(a) =0$ and $f'$ is bounded we get $|f(x)|\leq M|x-a| = M(x-a)$

Similarly for $x \in [\frac{a+b}{2}, b]$ we obtain $|f(x)|\leq M|x-b| = M(b-x)$ Note that both the above relations cannot be equalities simultaneously or else the right hand derivative of $f$ at $\frac{a+b}{2}$ would be $-M$ and the left hand derivative would be $M,$ a contradiction. The rest is straightforward:

$\begin{align} \displaystyle \int_{a}^{b}f(x) dx &= \int_{a}^{\frac{a+b}{2}}f(x) dx+ \int_{\frac{a+b}{2}}^{b}f(x)dx \\&< \int_{a}^{\frac{a+b}{2}}(x-a)dx + \int_{\frac{a+b}{2}}^{b}(b-x)dx\\&= M\dfrac{(b-a)^2}{4} \end{align}$

$\Rightarrow \dfrac{4}{(b-a)^2} \displaystyle\int_{a}^{b}f(x)dx <M$

Since $M = \sup\{|f'(x)|: x\in [a,b]\}, $ corresponding to $\varepsilon = M - \dfrac{4}{(b-a)^2} \displaystyle\int_{a}^{b}f(x)dx >0$ there exists $c \in [a , b]$ such that $|f'(c)| > M-\varepsilon = \dfrac{4}{(b-a)^2}\displaystyle\int_{a}^{b}f(x)dx, $ thus completing the proof.

For the problem in question, let $f(t)$ denote the velocity function as a function of time so that $f'(t)$ denotes acceleration as a function of time. Given: $f(0)=f(1)= 0$ and $\displaystyle \int_{0}^{1}f(x)dx =1.$ Then from the result above there exists $c \in [0, 1]$ such that $|f'(c)|>4.$

Note that for equality we would require the velocity function to not be differentiable at $0.5$, a case shown in the diagram posted by the OP which is a physical impossibility.