A particle which travels a unit distance in a unit time, and starts and ends with velocity 0, has at some time an acceleration $\ge 4$. (a) Prove that if $f$ is a twice differentiable function with $f(0) = 0$ and $f(1) = 1$ and $f'(0)=f'(1)=0$, then $|f''(x)| \ge 4$ for some $x \in (0,1)$. Hint: Prove that either $f''(x) \ge 4$ for some $x \in (0,1/2)$, or else $f''(x) \le -4$ for some $x \in (1/2,1)$. 
(b) Show that in fact we must have $|f''(x)| \gt 4$ for some $x \in (0,1)$. 
For (a), following the hint, first suppose that $f''(x) \lt 4$ for all $x \in [0,1/2]$. Then by the Mean Value Theorem, for all $x$ in $[0,1/2]$ we have 
$$\frac{f'(x)-f'(0)}{x-0} = f''(x') \lt 4$$ for some $x' \in [0,x]$ so $f'(x) \lt 4x$. If we set $h(x) = 2x^2$, then $f(0)=h(0)$ and $f'(x) \lt h'(x)$, and it follows that $f(x) \lt h(x)$ on $[0,1/2], so that $f(1/2) \lt 1/2.
Now if $f''(x) \lt -4$ on $[1/2,1]$ then by the same sort of analysis, $$\frac{f'(1)-f'(x)}{1-x} = f''(x') \lt 4$$ for some $x' \in [x,1]$, so $f'(x) \lt 4-4x$, then if we set $h(x)=4x-2x^2$, we cannot get a result like above. I'm assuming I need to lead to the result $f(1/2) \gt 1/2$ to derive a contradiction, but I don't see how. 
Can anyone explain how I can solve this problem and also (b), I've been stuck on this problem for some time now I'd greatly appreciate any help.
 A: You are on the right track. You already showed that $f(\frac12)<\frac12$ if $f''(x)<4$ for all $x\in(0,\frac12)$. For the right half, note that you should pick your $h$ such that $h(1)=f(1)$, that is $h(x)=4x-2x^2-1$ (and watch your signs and order relation directions!). Then the same calculation works. 
Alternatively, note that if $f$ has the properties of the problem statement, then so does $\tilde f(x)= 1-f(1-x)$. Then $f''(x)>-4$ for some $x\in(\frac12,1)$ translates to $\tilde f''(x)<4$ for some $x\in(0,\frac12)$; we know already that this  implies $f''(\frac12)<\frac12$, i.e., $f(\frac12)>\frac12$, the desired contradiction.
As for (b), revisit the above proof. If we only demand $|f''(x)|\le 4$, then we only get $f'(x)\le 4x$ for $x\in(0,\frac12)$ and correspondingly $f'(x)\ge -4x$ for $x\in(\frac12,1)$. As $f''$ exists, $f'$ must be continuous, hence the inequality must be strict in some open subinterval. In that interval, the comparison with $2x^2$ or $1-2(1-x)^2$ must be strict.
