Q:

Find piecewise constant function u for $X'(t)=AX(t) + Bu(t)$ and $X(t)=\begin{pmatrix}10 \\0 \end{pmatrix}$ for some T

Consider the system
$$x''(t)=u(t)$$
such that $x(0)=100, \; x'(0)=50$. Find a function $u$ piecewise constant such that $x(T)=0, \; x'(T)=10$ for a time $T$
Using the control theory language, it is asking for a piecewise constant admissible control such that the point $(0,10)$ is reached by $(100,50)$ in some time $T$.
My attempt 
I transformed this into the system
$$\begin{bmatrix} x' \\ y'\end{bmatrix} = \begin{bmatrix} 0 & 1 \\ 0 & 0\end{bmatrix}\begin{bmatrix} x \\ y\end{bmatrix}+\begin{bmatrix} 0 \\ 1\end{bmatrix}u$$
which has the solution
$$\begin{bmatrix} x \\ y\end{bmatrix} = \begin{bmatrix}1 & t \\ 0 & 1 \end{bmatrix}\begin{bmatrix}100 \\ 50  \end{bmatrix} + \int_0^t \begin{bmatrix} (t-s)u \\ u\end{bmatrix}ds$$
Solving the integral, considering $u$ a constant, I got that
$$\begin{bmatrix} x \\ y\end{bmatrix} = \begin{bmatrix} 100 + 50t + \frac{t^2}{2} u \\ 50 + tu\end{bmatrix}$$
I've tried to make this equal to the given vector that we want, but I couldnt find t and u that matches. What is wrong?
Thanks!
Edit
Now I see that I made a mistake calculating the solution, so I fixed the last matrix, made it equal to the vector that I wanted and solved for t and u. Answer can be found here

A:

Because this is a 2nd order system, you probably cannot find an $u$ that transfers the states to the desired value in 1 step, which you have failed to do. Since this system is reachable and 2nd order, you can reach any state at most 2 steps, i.e. using 2 constant values of $u$ (and I think this is implied by "piecewise constant"). In order to find such $u$, you need to discretize the system first as follows:
$$x((k+1) \mu) = F(\mu) x(k \mu) + G(\mu) u(k \mu)$$
where $\mu$ is the sampling period, $F(\mu) := e^{A \mu}$ and $G(\mu) := \left( \int_0^\mu e^{A \tau} d \tau \right) B$. This is standard literature. Now, observe that
$$x(T) = x(2 \mu) = F^2(\mu) x(0) + \begin{bmatrix}G(\mu) & F(\mu) G(\mu)\end{bmatrix} \begin{bmatrix}u(\mu) \\ u(0)\end{bmatrix}$$
Then, finding $u$ values is a standard linear equation problem. You can also select any $\mu$ as long as it is non-pathological, i.e. $\mu (\lambda - \gamma) \neq 2i\pi k, (k=\mp 1, \mp 2, \dots)$ where $\lambda$ and $\gamma$ are 2 eigenvalues of $A$.

