# Time Dependent Acceleration

1. Sep 4, 2011

### waters

The function ax(t) describes the acceleration of a particle moving along the x-axis. At time t = 0, the particle is located at the position x0 and the velocity of the particle is zero.

ax(t) = (a0)(e^(-bt))

The numerical values of all parameters are listed below:

x0 = 7.10 m
a0 = 9.00 m/s2
b = 8.00 s−1
t1 = 1.60 s
t2 = 1.40 s

A) Calculate the change in velocity ∆vx between time t=0 and t=t1.
The answer is 1.12 m/s. That was easy.

B) Calculate the change in position ∆x between time t=0 and t = t1.
The answer is NOT -.1406 m or 6.96 m. I integrated the acceleration function twice and got -.1406 (the change in displacement). How does that even work? How can your velocity be positive but your displacement negative during the same time period?

C) Calculate the average velocity vx, ave between time t=0 and t=t2.

D) Calculate the average acceleration of the particle ax, ave between time t=0 and t=t2.
The answer is .804 m/s^2; I did the same thing as above but with acceleration. Why does it work now?

Last edited: Sep 4, 2011
2. Sep 4, 2011

### cepheid

Staff Emeritus
Welcome to PF waters!

I disagree with your result for part A. Are you sure that you are integrating correctly? Can you post the steps of your integration?

3. Sep 4, 2011

### waters

Thanks for responding. I am hopelessly lost.

Sure.

http://www.wolframalpha.com/input/?i=integrate+9e^%28-8t%29+from+0+to+1.6

It rounds up to 1.125. Is there something that I'm doing wrong?

The parts I really need help with are B and C.

4. Sep 4, 2011

### cepheid

Staff Emeritus
Actually never mind. I had made a small arithmetic error. Now my answer matches yours. Let's move on to part B. Same question. Can you post the steps of your integration?

5. Sep 4, 2011

### waters

Sure. Sorry I probably should have done this all in my first post.

I'll post C right after this:

B)
http://www.wolframalpha.com/input/?i=integrate+%28-9%2F8%29e^%28-8t%29+from+0+to+1.4

6. Sep 4, 2011

### waters

For B, it says change in displacement, so I thought it would be -.14. 6.96 is the displacement itself (7.1 - .14), and I put that into the computer, and it was still incorrect.

C) The value of the integral for B multiplied by (1/1.4)

7. Sep 4, 2011

### cepheid

Staff Emeritus
Well, that explains it. You are integrating the wrong function. That expression for velocity as a function of time is incorrect. Wolfram Alpha is great for checking your answers and finding mistakes, but in solving the problem, you should actually integrate by hand and find analytical expressions for the functions vx(t) and x(t). Then plug in the numbers into those expressions. For instance, for part A, we have:

$$v_x(t_1) - v_x(0) = \int_0^{t_1} a_x(t)\,dt$$

However, vx(0) = v0 = 0, so we get rid of it. Now, plug in the expression for the acceleration and integrate:

$$v_x(t_1) = \int_0^{t_1} a_0 e^{-bt}\,dt$$

$$= a_0\left[-\frac{1}{b}e^{-bt}\right]_0^{t_1}$$

$$= \frac{a_0}{b}(1 - e^{-bt_1})$$

Now, for part B, you have to integrate this function i.e. $v_x(t) = (a_0/b)(1 - e^{-bt})$ with respect to time in order to get x(t). Can you take it from here?

8. Sep 4, 2011

### waters

Wow thanks; that was very helpful. Let me see if this sinks in a bit more and I will edit this post if I still don't understand something.

Again, thank you very much. I really do appreciate it.

Wait, I'm still getting the wrong answer. It's not 1.43 m

Last edited: Sep 4, 2011
9. Sep 4, 2011

### waters

I integrated (9/8) - (9/8)(e^-8t). I got (9/8) + (9/64)(e^-8t), and that evaluated from 0 to 1.4 is 1.43 m. This checks out on wolframalpha. What am I doing wrong?

The answer isn't even 8.53 m. I am so lost.

10. Sep 4, 2011

### cepheid

Staff Emeritus
Well, first of all, in part B, aren't you still supposed to consider things at t1, which is 1.6 s (not 1.4 s)?

But more importantly, it's pretty clear that your integration must be wrong, because you start with a term that is just 9/8, and you still end up with that 9/8 at the end. But the integral of a constant is not a constant. Here are the first few steps:

$$x(t) - x(0) = \frac{a_0}{b}\int_0^t (1 - e^{-b\tau})\,d\tau$$

x(0) = x0 (the initial position) and so:

$$x(t) = \frac{a_0}{b}\left[ \int_0^t \,d\tau - \int_0^t e^{-b\tau} \,d\tau \right] + x_0$$

What would you do next? By the way, notice that I kept the upper limit of integration as a variable, so that the answer is a function, rather than a number (it tells you what the position is up to time t, for any value of t). Notice that I used another symbol tau for the time variable inside the integral, in order to distinguish the two variables. Tau is what's called a dummy integration variable. Anyway, this is one way of getting an expression for x(t) as a general function of time. Another way is to compute the indefinite integral $x(t) = \int v_x(t)\,dt$. The two methods are equivalent, since in the case of the indefinite integral, you'll end up with a constant of integration C on the right hand side, and when you solve for C, you'll find that C = x0. The physical interpretation of constants of integration in situations like this is that they are initial values of a function, since the integral only gives you the change in value of the function, but you need to add the starting value to that to get the present value at time t. Am I making any sense?

11. Sep 4, 2011

### waters

Never mind; I had the wrong limits of integration. It works fine. Thanks. Sorry for the confusion.

Just one question: The way I attempted to do the problem in the first place, why was it so wrong? Why can't you directly integrate the integral of the integral of acceleration? Why do you have to evaluate the integral of acceleration using the limits of integration, and then evaluate that expression for displacement?

I haven't done calculus in such a long time, so my calculus skills are rusty. Sorry about that.

Yes, you are making sense. I understand.

Last edited: Sep 4, 2011
12. Sep 4, 2011

### cepheid

Staff Emeritus
There was nothing wrong with your method. Acceleration is the second derivative of position with respect to time. So to get position vs. time given acceleration, you'd integrate the function a(t) twice. It's just that you weren't careful and didn't do it by hand, so the first integral resulted in a function v(t) that you didn't record anywhere (since a computer did it for you), and for the second integral, you didn't integrate the result of the first integral. All I've been doing for you is going through the steps of those two integrations carefully.

13. Sep 4, 2011

### cepheid

Staff Emeritus
You keep editing your post. Anyway, as I explained in post #10, there's no reason why you can't do it using indefinite integrals rather than definite integrals. Using indefinite integrals it would be:

$$v(t) = \int a(t)\,dt$$

$$x(t) = \int v(t)\,dt = \int\left(\int a(t)\,dt\right)\,dt$$

You just have to be careful and keep track of all of the constants/initial values.

14. Sep 4, 2011

### cepheid

Staff Emeritus
For instance, in this example:

$$v(t) = \int a(t)\,dt = a_0 \int e^{-bt}\,dt = -\frac{a_0}{b}e^{-bt} + C$$

where C is the constant of integration. To solve for C, set t = 0 to obtain:

$$v(0) = -\frac{a_0}{b}e^{0} + C = -\frac{a_0}{b} + C$$

Using the initial condition you were given, that v(0) = 0, we end up with C = a0/b and the function becomes:

$$v(t) = \frac{a_0}{b} -\frac{a_0}{b}e^{-bt}$$

To get x(t), you have to integrate this, and that integration will have its own constant of integratio, which you will have to solve for using the initial condition for x. Without the initial condition, there is no unique solution, since a given function has a whole family of antiderivatives (EDIT: this is equivalent to saying that after integrating, you know the change in the function over a time interval, but you don't know the starting value, so your result could be off by any additive constant).

If you use definite integrals (i.e. explicitly include the limits of integration to define the time interval) then you don't have to be so meticulous about the constants, since they appear automatically in the final answer (as a result of evaluating the integral at the lower endpoint of the interval). Either method is fine though.

Last edited: Sep 4, 2011
15. Sep 4, 2011

### waters

Okay, I see. I know that it gets annoying if I keep editing my post, but I was still trying to digest everything. But thanks for noticing.

I don't think I have anymore questions.