Interesting math question motivated by physics -- if $\int_0^1f(x)\,dx=1$ and f(0)=f(1)=0 then find $\min f'(x)$ I saw a question in a physics textbook which was:

A particle starts from rest at $t=0$, $x=0$, and comes again at rest on $x=1$, $t=1$. Let the instantaneous acceleration be $a$. Then:
(a) $a$ cannot be positive for all $t\in[0,1]$.
(b) $|a|$ cannot exceed 2 at any point.
(c) $|a|$ must be $\geq 4$ at some point
(d) $a$ must change sign but no other assertion can be made.

The anwer was (a,c) which is easy to see intuitively but how to attack this problem rigorously?
Translated into math the problem becomes

$$\text{If }\int_0^1 f(x) \, dx = 1,\text{ and }f(0)=f(1)=0\text{ then find } \min f'(x).$$
Assuming $f(x)$ is continuous.

At first I thought lets use some integral inequalities like Cauchy-Schwarz or something but then I realised it wont work. After that I thought that this is somewhat similar to calculus of variations but not quite. I have no idea how to approach this rigorously but I want to see how its done because this is like a "simple problem" but still I have no clue how to approach this. ... Its probably indicating some sort of gap in my knowledge or technique. So .. Please help!
 A: If we denote the distance at time $t$ with $s(t)$ then
$$
 s(0) = 0 \, , \quad s(1) = 1
$$
and
$$
 s'(0) = s'(1) = 0 \, .
$$
$s$ is the antiderivative of your function $f$, and $s''(t)$ is the
acceleration at time $t$.
The mean-value theorem applied to $s'$ implies that $s''(t_0) = 0$
for some $t_0 \in (0, 1)$, which is statement (a).
(Since $s'$ cannot be constant, one can even conclude that $s''$
must be strictly negative at some point.)
For (c) we can use Taylor's theorem with the explicit formulas for the remainder. First on the interval $[0, \frac 12]$:
$$
 s(\frac 12) = s(0) + s'(0) \frac 12 + \frac{s''(t_1)}{2}\left( \frac 12 \right)^2 = \frac 18 s''(t_1)
$$
for some $t_1 \in (0, \frac 12)$, and then on the interval $[\frac 12, 1]$:
$$
 s(\frac 12) = s(1) + s'(1) (1-\frac 12) + \frac{s''(t_2)}{2}\left( 1-\frac 12 \right)^2 = 1 + \frac 18 s''(t_2)
$$
for some $t_2 \in (\frac 12, 1)$. It follows that
$\frac 18 s''(t_1) = 1 + \frac 18 s''(t_2)$, or
$$
8 = s''(t_1) -  s''(t_2) \le \lvert s''(t_1) \rvert + \lvert s''(t_2) \rvert
$$
and therefore (at least) one of $\lvert s''(t_1) \rvert$, $\lvert s''(t_2) \rvert$ must be $\ge 4$.
A: As stated, the answer to your problem of finding the min value of $f'(x)$ is $-\infty$. Your problem should really be phrased as finding the largest $c$ such that $\sup_{t \in [0,1]} |f'(t)| \geq c$ is guaranteed. 
The above answer by Martin shows that $c=4$ works. Here is a different approach to showing that $c=4$ works. It further shows we cannot increase $c$ beyond 4.

Define the continuous "tent" function $w(t)$:
$$ w(t) =   \left\{ \begin{array}{ll}
4t &\mbox{ if $t \in [0,1/2]$} \\
4-4t  & \mbox{ if $t \in [1/2, 1]$} 
\end{array}
\right.$$
Notice that $w(0)=w(1)=0$, $\int_0^1 w(t) dt = 1$,  and  $|w'(t)|=4$ whenever $t \neq 1/2$. Now suppose we have a differentiable function $f$ such that $f(0)=f(1)=0$ and $\int_0^1 f(t)dt = 1$.  
a)  Show that $f(t) < w(t)$ for all $t \in (0,1)$ is impossible, and $f(t)>w(t)$ for all $t \in (0,1)$ is also impossible.  Hence, the two continuous curves $f$ and $w$ must touch each other at some point in $(0,1)$. [Hint: Compare integrals of these continuous functions.]
b) From (a) we know there is a point $x \in (0,1)$ such that $f(x)=w(x)$. Use the mean value theorem (as suggested by amd comments) to show that there is a point $v \in (0,1)$ such that $|f'(v)|=4$.
c) Fix $\epsilon>0$.  Modify $w(t)$ to produce a new curve $\tilde{w}(t)$ that is differentiable and satisfies $\tilde{w}(0)=\tilde{w}(1)=0$, $\int_0^1\tilde{w}(t)dt=1$, $|w'(t)|\leq 4+\epsilon$ for all $t \in [0,1]$. [Hint: Slightly increase the slope of $w(t)$ and round out the middle so it is differentiable.]  
