How to use a piecewise acceleration function to get a position function?

Question

This should be a relatively easy problem but I think I am missing something somewhere. This problem consists of a object that is being thrown into the air at $t = 4s$ at a velocity $v_0$

here is my acceleration function: $a(n) = \begin{cases} 0, & t<t_1 \\ g, & t≥t_1 \end{cases} $

Where

$g = - 9.8m/s^2$

$t_1 = 4s$

$x_0=0m$

$v_0$ is the velocity at which the object is being thrown up in the air at.

When I derive the velocity function it seems to be correct from what I could find, $v(n) = \begin{cases} 0, & t<t_1 \\ gt-gt_1+v_0, & t≥t_1 \end{cases} $

But when I go to derive the position function I get lost.

$y(t) - x_0= \int_{t_1}^tv(t)dt => [\frac{1}{2}gt^2-gt{t_1}+v_ot]_{t_1}^t =>\frac{1}{2}gt^2-gt{t_1}+v_ot -(\frac{1}{2}g{t_1}^2-g{t_1}^2+v_ot_1) $

When I then go apply this to the rest of the problem I get nonsense answers.

Can someone please let me know where I've gone wrong. Sorry if this is an easy problem, I am a beginner to physics.

PS: I know you can solve for this algebraically, you get $y(t) = \begin{cases} 0, & t>t_1 \\ \frac{1}{2}g(t+t_1)+v_0(t+t_1), & t≥t_1 \end{cases}$ but I would like to know the derivation based on the calculus of the problem as it is more relevant to the course I am following.

Answer 1

Your integration is correct. It just needs some completing-the-square TLC.

$$ \frac{1}{2} g t^2 - g t t_1 + v_0 t - \frac{1}{2} g t_1^2 + g t_1^2 - v_0 t_1 $$

$$ \frac{1}{2} g\left( t^2 - 2 t t_1 + t_1^2 \right) + v_0 (t-t_1) $$

$$ \frac{1}{2} g\left(t - t_1 \right)^2 + v_0 (t - t_1) $$

which is what your PS should have said.

Notice that we can, from the very beginning of the problem, make a change of variables

$$ t' = t - t_1 $$

Then all integrals are from $t' = 0$ and we find

$$v(t') = gt' + v_0$$ and $$x(t') = \frac{1}{2}g(t')^2 + v_0 t' $$