# Displacement from Acceleration/Time graph

1. Jul 2, 2017

### Raymour

1. The problem statement, all variables and given/known data

2. Relevant equations
I know that I probably need to use s=vit+1/2at2, however, I'm having trouble properly applying it

3. The attempt at a solution
I know that part A is 20m/s and then 5m/s. I know this is an easy problem and I'm actually 2 chapters ahead. For some reason, I am struggling with this! The book is telling me the answer should be 263 m.

Last edited: Jul 2, 2017
2. Jul 2, 2017

### phinds

Rather than an image, I just see the text

3. Jul 2, 2017

### Raymour

I think I fixed it!

4. Jul 2, 2017

### phinds

--- OOPS --- post deleted. See post #8

Last edited: Jul 2, 2017
5. Jul 2, 2017

### Raymour

Shouldn't the change in velocity be acceleration x time? Starting from rest a change in velocity of 2 m/s/s for 10 s should result in a speed of 20m/s? That is the answer my book gives as well...

6. Jul 2, 2017

### Ray Vickson

No: change in velocity = integral of acceleration over time. This gives time x acceleration only when acceleration is constant.

7. Jul 2, 2017

### Raymour

From 0 to 10 seconds in this problem, the acceleration is constant. If not 20 m/s what should it be? That's why I'm struggling with part B as I'm not sure what to integrate to get my distance traveled (Not displacement sorry!).

Last edited: Jul 2, 2017
8. Jul 2, 2017

### phinds

Sorry, I was getting distance confused w/ velocity. I agree w/ you on the velocity after 10 seconds being 20m/s

9. Jul 2, 2017

### Ray Vickson

If course the velocity at t = 10 sec is 20 m/s; nobody claimed otherwise. Remember, to do the whole question you also need the position at that time.

So you know the velocity and position at t = 10 - 0.0000000001 sec. Do you think the velocity and position will suddenly jump to new values just at 10 sec, or can you say also what the velocity and position are at t = 10 + 0.0000000001 sec? From that, can you get the velocity and position at t = 15 sec? Continue like that.

10. Jul 2, 2017

### phinds

Well, actually, at one point I did, but that's only because I'm stupid and I corrected myself.

11. Jul 2, 2017

### Raymour

I'm sorry, I'm seriously inept at Physics!! I have't taken it since high school 3 years ago and jumped into an independent study Calc-based... not my smartest decision. I'm not entirely sure what you mean by that.

12. Jul 2, 2017

### Ray Vickson

Mean by what, exactly?

I asked you whether you think the particle (or whatever it is) can suddenly change its velocity at a point in time. Would that not require an infinite acceleration? Do you think that a particle can suddenly jump (instantly) from one place to another? Would that not require infinite speed?

Anyway, the problem does not require any physics, only elementary calculus. Are you saying you have never taken calculus, or that you have forgotten it? If so, we may be able to suggest readings and on-line tutorials and the like, but in any case you would really not be ready to tackle a problem like this one----you would need to learn and absorb a number of pretty basic concepts first.

13. Jul 2, 2017

### jack action

This is the right equation for a situation where the acceleration is constant. If you look at the problem carefully, you'll notice that it can be separated in 3 parts where the acceleration is constant.

14. Jul 2, 2017

### Raymour

I'm currently taking differencial equations and have passed Calculus I,II, and III all with an A... it's more so that I'm not familiar with applying Calculus to these problems (nor do I have a lecture in order to show me and my book seems to be useless as the chapter in which this problem is does not give such an example)... Am I integrating (twice of course to velocity then to position) with respect to time at 3 seperate points in time, corresponding to the 3 different accelerations? None of the answers I have come up with have matched my book's answer key so I'm not sure if I'm simply not using the correct numbers. Or do I integrate the velocity with respect to time (20m/s from 0-15s and 5 m/s from 15-20s)? I feel like there's something that's common sense that I am just missing.

Last edited: Jul 2, 2017
15. Jul 2, 2017

### Raymour

Ahh I got it! Thank you! I tried that earlier, however, instead of using 10s, 5s, and 5s I was using the fully elapsed time... whoops! I would still really like to know how to properly solve this with integration, however.

16. Jul 2, 2017

### LCKurtz

OK, since you have figured out the answer I think I can get away with showing you how to properly set it up. You are given$$a(t)=\begin{cases} 2,&t<10\\ 0, & 10 < t < 15\\ -3, & 15 < t < 20 \end{cases}$$with initial position and velocity $0$. Now, to get the velocity at time $t$ you need to do the integral$$v(t) = \int_0^t a(u)~du,~0\le t\le 20$$Evaluating this integral depends on which interval $t$ is located. For example, for $0\le t \le 10$ you just have$$v(t) = \int_0^t a(u)~du = \int_0^t 2~du = 2t$$Now, if $10 \le t \le 15$ your integral becomes$$v(t) =\int_0^t a(u)~du = \int_0^{10} a(u)~du + \int_{10}^t a(u)~du =\int_0^{10} 2~du + \int_{10}^t 0~du = 20$$Remember, this is just for $10\le t \le 15$. Finally, if $15\le t \le 20$, you have$$v(t) =\int_0^t a(u)~du = \int_0^{10} a(u)~du + \int_{10}^{15} a(u)~du + \int_{15}^{t} a(u)~du$$ $$= \int_0^{10} 2~du + \int_{10}^{15} 0~du + \int_{15}^{t} -3~du = 20 + 0 -3(t-15)$$Remember this last one is just for $15\le t \le 20$. So putting this all together you have the three piece formula for velocity:$$v(t) = \begin{cases} 2t, & 0\le t \le 10\\ 20, & 10\le t \le 15\\ 20 -3(t-15), & 15\le t \le 20 \end{cases}$$In the above I have used the inital zero position and velocity without explicitly pointing it out. I have added extra steps for clarity, but if you understand this, you can do the same thing to calculate the position $s(t)$ using the three piece formula we have for $v(t)$. Try it.

Last edited: Jul 2, 2017
17. Jul 2, 2017

### jack action

Here what would be my method.

Integrating a constant acceleration to find velocity:
$$dv = adt$$
$$\int_{v_0}^{v}dv = \int_{t_0}^{t}adt$$
$$v - v_0 = a (t - t_0)$$
So you have an equation for $v$:
$$v = v_0 + a (t - t_0)$$
Integrating velocity to find the distance:
$$dx = vdt$$
$$\int_{x_0}^{x}dx = \int_{t_0}^{t}(v_0 + at - at_0)dt$$
$$x - x_0 = v_0 t + \frac{1}{2}at^2 - at_0 t - (v_0 t_0 + \frac{1}{2}at_0^2 - at_0 t_0)$$
So you have an equation for the distance traveled $x - x_0$:
$$x - x_0 = (v_0 - at_0) (t - t_0) + \frac{1}{2}a(t^2 - t_0^2)$$
With those equations, you can use them with the actual numbers on the graph for the time (0 &10, 10 & 15, 15 & 20).

Now, the distance traveled for $t = 0$ to $t = 20$ is:
$$\int_{x_0}^{x_{20}}dx = \int_{x_0}^{x_{10}}dx + \int_{x_{10}}^{x_{15}}dx + \int_{x_{15}}^{x_{20}}dx$$
We already know the solution of $\int dx$ when the acceleration is constant for each part, so:
\begin{aligned} x_{20} - x_0 =& (v_0 - (2) t_0) (t_{10} - t_0) + \frac{1}{2}(2)(t_{10}^2 - t_0^2)\\ &+ (v_{10} - (0) t_{10}) (t_{15} - t_{10}) + \frac{1}{2}(0)(t_{15}^2 - t_{10}^2)\\ &+ (v_{15} - (-3) t_{15}) (t_{20} - t_{15}) + \frac{1}{2}(-3)(t_{20}^2 - t_{15}^2)\\ x_{20} - x_0 =& ((0) - (2)(0)) ((10) - (0)) + \frac{1}{2}(2)((10)^2 - (0)^2)\\ &+ ((20) - (0)(10)) ((15) - (10)) + \frac{1}{2}(0)((15)^2 - (10)^2)\\ &+ ((20) - (-3)(15)) ((20) - (15)) + \frac{1}{2}(-3)((20)^2 - (15)^2)\\ x_{20} - x_0 =&\ 4 + 100 + 62.5 \\ x_{20} - x_0 =&\ 166.5 \end{aligned}

18. Jul 2, 2017

### Raymour

Ahh thank you so much!! I was trying to simply integrate at 10, 15, and 20s instead of adding each preceding integral of the time intervals into the 2nd and 3rd acceleration integrations.

19. Jul 2, 2017

### Raymour

Thank you! Now that I see it applied, I completely understand. I was struggling to see that I must integrate each interval of time that the acceleration is constant and not just the time at a specific point when the acceleration changes. I really like your method of integrating the equations first and then putting in the numbers so I could see exactly what was happening. Thanks so much! I probably should have taken this class concurrently with Calc II but I didn't think I would need it as a biologist... graduate school had a different idea lol!

20. Jul 2, 2017

### Ray Vickson

Well, had I realized you know calculus, etc., I would not have needed to be so careful in my explanations. I could just say that the position and velocity $x(t)$ and $v(t) = dx(t)/dt$ are continuous functions of $t$, so the only issue is how to stitch together the three constant-acceleration pieces to make things come out right. The point I was trying to make was that the end conditions for one interval become the initial conditions for the next interval. That comes out directly from the properties of integration itself---to get $v(t)$ you are just integrating a "step function" (giving a continuous function as the answer) and to get $x(t)$ you are integrating the continuous function $v(\cdot)$ Of course, the integrations have already been done when you look at the constant-acceleration results $v_i(t) = v_{i0} + a_i (t-t_i)$ and $x_i(t) = x_{i0} + v_{i0} (t-t_i) + (1/2) a_i (t-t_i)^2$ for $t$ in the interval $I_i = (t_i, t_{i+1})$.