Related rate problem (absolute value of its acceleration should be at least 4.) If a particle moving on the Euclidean line traverses distance 1 in time
1 starting and ending at rest, then at some time t ∈ [0, 1], the absolute
value of its acceleration should be at least 4.
So,the distance function f(t) is continous,differentiable ,and f(0)=0,f(1)=1.
And  $\cfrac{d}{dt}f(0)=0$ and  $\cfrac{d}{dt}f(1)=0$.
I need to use calculus(whatever tools maybe) to solve this,
I'm confused , please give me a hint to proceed !
 A: There should be a more elegant solution, but I guess that this works as well.
Since $f$ is continuous, $f(0)=0$, and $f(1)=1$, the Intermediate value theorem gives us a $t_0 \in [0,1]$ such that $f(t_0) = \frac{1}{2}.$
Now we consider two cases, namely $t_0 \leq \frac{1}{2}$ and $t_0 \geq \frac{1}{2}.$ (I will only treat the former, since the argument for the latter is almost identical.)
The Mean value theorem provides us with a $t_1 \in [0, t_0],$ so that
$$ \vert f'(t_1) \vert = \left\vert \frac{f(t_0)-f(0)}{t_0-0} \right\vert
= \frac{\vert 1/2 - 0 \vert}{|t_0|} \geq \frac{1/2}{1/2} = 1. $$
Now we split into two cases again, namely $t_1 \leq \frac{t_0}{2}$ and $t_1 \geq \frac{t_0}{2}$. (I will again only treat the former.)
Utilizing the Mean value theorem one more time, we find a $t_2 \in [0,t_1],$ so that
$$ \vert f''(t_2) \vert = \left\vert \frac{f'(t_1)-f'(0)}{t_1-0} \right\vert
= \frac{\vert 1 - 0 \vert}{\vert t_1 \vert} \geq \frac{1}{|t_0|/2} = \frac{2}{t_0} \geq \frac{2}{1/2} = 4,$$
which was to be proven.
I leave the rest of the cases as an exercise. The only difference should be how you set up your intervals for the Mean value theorem.
