$\int 0.733xe^{-0.27x}$

Question

this is my first time using this website, so pls tell me if I didn't follow any rules. I'm doing a math investigation, which I'm trying to find the displacement function based on the velocity function using integration. $$V(t)=0.733xe^{-0.27x},$$ I got $$S(t)=10.05(-0.27e^{-0.27x}x-e^{-0.27x}).$$ However, the graph doesn't make sense. The velocity is positive when t is greater than 0, but the displacement function is negative. Could anyone help me with this?

Answer 1

In general, if $s(t) = \int v(t) \, \mathrm{d}t$ is an antiderivative of a velocity function $v(t)$, then since $v(t) = \frac{\mathrm{d}}{\mathrm{d}t}x(t)$ where $x$ is the position (or displacement) function, we get that $$x(t) = \int \frac{\mathrm{d}}{\mathrm{d}t}x(t)\, \mathrm{d}t = \int v(t) \, \mathrm{d}t$$ so $x(t)$ is also an antiderivative of the velocity. Recalling that two antiderivatives differ by a constant, we can say that $$ x(t) = s(t)+c\tag{1} $$ where $c$ is some constant. But what exactly is that constant $c$? Well, if we evaluate $(1)$ at time $t=0$, then by definition $x(0)=x_0$ is the position of the moving object at the start of its trajectory. We thus get $$ c = x_0 -s(0) \color{blue}{\implies} \boxed{x(t) = s(t) -s(0)+x_0 }\tag{2} $$

---

In your case, you calculated one possible antiderivative $s(t)$, but to find the displacement function you need to add the respective constants necessary as noted in $(2)$. Thus the correct displacement function is $$ x(t) = s(t)-s_0 +x_0 = \underbrace{e^{-0.27 x} (-2.71481 x - 10.0549)}_{s(t)} -\underbrace{(-10.0549)}_{s(0)} +x_0 $$ Thus, the position function you want is $$ \boxed{x(t) = 10.0549 e^{-0.27 x} (-1 + e^{0.27 x} - 0.269999 x) +x_0} $$ where $x_0$ is whatever initial position the particle starts at. Here's a graph of what the previous function looks like, where you can edit the value of $x_0$.

---

All this said, you mention the following in the body of your question:

However, the graph doesn't make sense. The velocity is positive when t is greater than 0, but the displacement function is negative.

This isn't an issue, we just need to think about how this translates into the real object you're trying to model.

Let's say the velocity is of a runner on a straight racetrack. Let's say that the "usual" place the runner starts running from is on a starting line marked on the track, i.e, we'll say the displacement is $0$ at the starting line.

So now let's suppose the runner starts his run behind the starting line. Since at the starting line we have displacement $0$, if the runner starts behind the starting line his initial position will be negative since its behind $0$. This scenario would correspond to the graph you got.

Alternatively, if the runner starts the run after the starting line, then his initial position will be positive. If he starts exactly at the starting line, then his initial position is $0$. The key thing here is that the runner will run exactly the same independently of where he starts running from, so the shape of his displacement graph will be the same, it will just vary where it starts from.

---

tl;dr: You forgot to add the constant, which in your problem just means choosing if your runner starts before, at or after the starting line; but all options make physical sense. Mathematically the constant "moves" the shape of the graph as you can see for yourself here.