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

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

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

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.

• If $v(t) = 0$ for $t < t_1$ and then $v(t_1) = v_0 > 0$ starting from $t = t_1$ then there is a discontinuity in $v$ at that time, are you sure this is what you mean? It does not follow from the given $a(t)$ at least, and seems unpysical. Feb 15, 2022 at 21:41
• Also, if by your PS you mean that $x(t)$ is supposed to be $\frac{1}{2}g(t+t_1) + v_0 (t+t_1)$ then that is also wrong, for starters the expression does not have the correct dimensions (the first term has units of speed). Feb 15, 2022 at 21:44
• I made a few quick edits to hopefully clear things up. I should mention that the equation in my PS does in fact work for the problem in terms of the numerical value it spits out, though I do see your point about the dimensions. Feb 15, 2022 at 21:53
• "(...)the equation in my PS does in fact work for the problem in terms of the numerical value it spits out(...)" is surely a coincidence. The numerical values don't mean anything without their units: if you have a question whose answer is supposed to be 100m/s, then the formula $50 kg + 50m/s$ does not achieve that result ;) Feb 15, 2022 at 21:59

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

• Thank you for your help! Feb 15, 2022 at 22:13
• You're welcome! Feb 15, 2022 at 22:18