Prove that if a particle travels a unit of distance in one unit of time starting and finishing in repose it has in a moment an acceleration $\ge 4$

Question

How would you solve this problem?

Prove that if a particle travels a unit of distance in one unit of time starting and ending whith velocity $0$ it has in a moment an acceleration $\ge 4$ (positive or negative).

I just know i have to start supposing that $-4 \lt a(t) \lt 4$ and using that $a(t)$ is the second derivate.

I am not allowed to use integration, just differential calculus.

Answer 1

Work from this:

If a particle starts at rest and the acceleration is $a = +4$ for $0 \le t \le 1/2$ and $a = -4$ for $1/2 \le t \le 1$, then the particle ends at rest at $x = 1$ at $t = 1$. If the acceleration is ever $a(t) < 4$ during the first half the particle will not arrive at the midpoint ($x = 1/2$) by $t = 1/2$.

Answer 2

Proof by Contradiction

Suppose $\frac{dv}{dt}=a<4$ for all times $t$ belonging to the interval $[0,1]$.

We then have $v(t)<4t+C$.

Noting that $v(1)=0$,

$0<4+C$ $\Rightarrow$ $C>-4$.

Hence:

$v(t)<4t-4$, which leads to,

$x(t)<2t^{2}-4t+B$.

But $x(1)=1$, therefore:

$1<2-4+B$ $\Rightarrow$ $B>3$.

And so we have:

$x(t)<2t^{2}-4t+3$

But remember that $x(1)=1$, and $2(1^{2})-4(1)+3=1$, hence our assumption was false.

There exists some time $t$ in the interval $[0,1]$ such that the acceleration is at least $4$.