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

Question

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.

Answer 1

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.

Answer 2

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