Question:
A particle moves in a straight line such that its velocity, $v$ in m/s, at time $t$ seconds, is given by:
$$v(t)=\left\{\begin{array}{cc} 5-(t-2)^2, & 0\leq t\leq 4\\ 3-\frac{t}{2},& t>4\end{array}\right.$$
the particle returns to its initial position at $t=T$. Find the value of $T$.
Doubt/Problem:
The mark scheme suggests that the initial position is equal to the distance travelled before coming to rest. Whereas I believe the initial position is the displacement at $t=0$, which I found out to be $4.33$ meters.
I am unable to identify what's wrong with my thinking pattern.
Mark scheme:
Let's call $x(t)$ to the particle position at time $t$. From the definition of velocity, $$v(t)=\frac{\mathrm{d}x}{\mathrm{d}t}\Longrightarrow\mathrm{d}x=v(t)\mathrm{d}t\Longrightarrow x(t)-x(0)=\int_0^tv(s)\mathrm{d}s\Longrightarrow x(t)=x(0)+\int_0^tv(s)\mathrm{d}s.$$
If $0\leq t\leq4$, the position is \begin{align*} x(t) & =x(0)+\int_0^t\bigl(5-(s-2)^2\bigr)\mathrm{d}s=x(0)+\int_0^t(-s^2+4s+1)\mathrm{d}s=\\ & =x(0)+\left[-\frac{s^3}{3}+2s^2+s\right]_0^t=x(0)-\frac{t^3}{3}+2t^2+t. \end{align*}
And, if $t>4$, \begin{align*} x(t) & =x(0)+\int_0^4\bigl(5-(s-2)^2)\mathrm{d}s+\int_4^t\left(3-\frac{s}{2}\right)\mathrm{d}s=\\ & =x(0)-\frac{4^3}{3}+2\cdot4^2+4+\left[3s-\frac{s^2}{4}\right]_4^t=\\ & =x(0)+\frac{44}{3}+3t-\frac{t^2}{4}-\left(12-\frac{16}{4}\right)=\\ & =x(0)+\frac{t^2}{4}+3t+\frac{20}{3}. \end{align*}
Now, let's see if the particle can get back to its initial position at $t\leq4$: $$x(0)-\frac{t^3}{3}+2t^2+t=x(0)\Longrightarrow t\left(-\frac{t^2}{2}+2t+1\right)=0.$$ The solutions of the above equation are $t_1=0$, $t_2=2+\sqrt{6}>4$ and $t_3=2-\sqrt{6}<0$ and none of them is valid for this case.
So, now we look for $t>4$. We have $$x(0)+\frac{t^2}{4}+3t+\frac{20}{3}=x(0)\Longrightarrow\frac{t^2}{4}+3t+\frac{20}{3}=0.$$ The solutions are $$t_4=6-2\sqrt{\frac{47}{3}}<0,\quad t_5=6+2\sqrt{\frac{47}{3}}\approx13.9.$$ So $T=t_5\approx 13.9$.