What is the distance a particle has moved if I know the work done


by Psinter
Tags: distance, moved, particle, work
Psinter
Psinter is offline
#1
Feb26-12, 12:54 PM
Psinter's Avatar
P: 72
1. The problem statement, all variables and given/known data
I was asked to find the distance a particle had traveled and was given its integral solved, except the upper limit was an equation.

[itex](x^2+2x)[/itex] is the force in pounds that acts on the particle


2. Relevant equations
[itex]\int_{0}^{x^2+2}(x^2+2x)dx = 108[/itex]


3. The attempt at a solution
I know I must find what number the top limit equation equals, but I've been inserting random numbers with no success. Also, I don't know if it is possible to find the value of x and plug it in. This is ridiculous, why couldn't they just give me the integral with the limits and I solve it.
Phys.Org News Partner Science news on Phys.org
Cougars' diverse diet helped them survive the Pleistocene mass extinction
Cyber risks can cause disruption on scale of 2008 crisis, study says
Mantis shrimp stronger than airplanes
alanlu
alanlu is offline
#2
Feb26-12, 12:58 PM
P: 65
What is the integral of x2 + 2x?
Psinter
Psinter is offline
#3
Feb26-12, 01:19 PM
Psinter's Avatar
P: 72
Quote Quote by alanlu View Post
What is the integral of x2 + 2x?
Its [itex]\frac{1}{3}x^3+x^2[/itex]

Ray Vickson
Ray Vickson is offline
#4
Feb26-12, 02:01 PM
HW Helper
Thanks
P: 4,672

What is the distance a particle has moved if I know the work done


Quote Quote by Psinter View Post
1. The problem statement, all variables and given/known data
I was asked to find the distance a particle had traveled and was given its integral solved, except the upper limit was an equation.

[itex](x^2+2x)[/itex] is the force in pounds that acts on the particle


2. Relevant equations
[itex]\int_{0}^{x^2+2}(x^2+2x)dx = 108[/itex]


3. The attempt at a solution
I know I must find what number the top limit equation equals, but I've been inserting random numbers with no success. Also, I don't know if it is possible to find the value of x and plug it in. This is ridiculous, why couldn't they just give me the integral with the limits and I solve it.
Maybe the problem is poor notation; this problem's notation is terrible--it double-uses the symbol x. Much better: write [tex] \int_{0}^{x^2+2}(x'^2+2x')dx' = 108,[/tex] which makes it clear what is the integration variable (x') and what is the unknown (x).

RGV
Psinter
Psinter is offline
#5
Feb26-12, 04:59 PM
Psinter's Avatar
P: 72
Quote Quote by Ray Vickson View Post
Maybe the problem is poor notation; this problem's notation is terrible--it double-uses the symbol x. Much better: write [tex] \int_{0}^{x^2+2}(x'^2+2x')dx' = 108,[/tex] which makes it clear what is the integration variable (x') and what is the unknown (x).

RGV
I see what you did there, but how does that makes it easier to solve? I still don't know what to do.

Rewinding everything I know:

The force exerted on the particle: [tex](x'^2+2x')[/tex]

The initial point of the particle: 0

The distance traveled by the particle (that's my upper limit) given by: [tex](x^2+2)[/tex]

And the total work done by the particle after traveling the distance I'm looking for: 108

Still not helps writing all that. How can I find the upper limit value or the x value to plug it in and get 108 as answer?
alanlu
alanlu is offline
#6
Feb26-12, 05:38 PM
P: 65
I agree with RGV. It may be easier to think about using a different variable name for the integral than the x you're trying to solve for.

Try plugging in the limits of integration as if you are evaluating the integral.
Psinter
Psinter is offline
#7
Feb26-12, 05:57 PM
Psinter's Avatar
P: 72
Quote Quote by alanlu View Post
I agree with RGV. It may be easier to think about using a different variable name for the integral than the x you're trying to solve for.

Try plugging in the limits of integration as if you are evaluating the integral.
I kept trying putting in random values and found that the upper value must be 6. In other words:
[tex] \int_{0}^{6}(x'^2+2x')dx' = 108,[/tex]
However, isn't there a process to reach that? I did it by brute force.
Ray Vickson
Ray Vickson is offline
#8
Feb26-12, 07:54 PM
HW Helper
Thanks
P: 4,672
Quote Quote by Psinter View Post
I kept trying putting in random values and found that the upper value must be 6. In other words:
[tex] \int_{0}^{6}(x'^2+2x')dx' = 108,[/tex]
However, isn't there a process to reach that? I did it by brute force.
You have already evaluated the indefinite integral [itex] F(x') = \int(x'^2 + 2x')\, dx'.[/itex] Using your formula for F, what would be [itex] \int_0^a (x'^2 + 2x') \,dx'[/itex]? What is preventing you from substituting in [itex] a = x^2 + 2[/itex]? Mind you, the final result will not be pretty, but it gives you a place to start.

RGV
Drood
Drood is offline
#9
Feb26-12, 08:10 PM
P: 8
Here's what I did, perhaps you may find the process of some use.

1) I computed the integral and applied the limits.
2) I know that the integral with the limit substitutions = 108
3) Now it's mostly a matter of expanding the terms and factoring to find solutions.
4) The only applicable solution I found was x=2
5) The upper limit with the x=2 substitution is indeed equal to 6.

Sorry for the short reply, I did it by hand and did the expansions and factoring by calculator since they were quite long. If you would like, I could write it out or post a picture of my solution if you did not understand my steps above.

Hope that helps,

Drood
Psinter
Psinter is offline
#10
Feb26-12, 08:50 PM
Psinter's Avatar
P: 72
Quote Quote by Ray Vickson View Post
You have already evaluated the indefinite integral [itex] F(x') = \int(x'^2 + 2x')\, dx'.[/itex] Using your formula for F, what would be [itex] \int_0^a (x'^2 + 2x') \,dx'[/itex]? What is preventing you from substituting in [itex] a = x^2 + 2[/itex]? Mind you, the final result will not be pretty, but it gives you a place to start.

RGV
It gives me x = 2, but I wonder if there is a rule or something for me to find that the value was 6 instead of doing it by guessing like I did.

Quote Quote by Drood View Post
Here's what I did, perhaps you may find the process of some use.

1) I computed the integral and applied the limits.
2) I know that the integral with the limit substitutions = 108
3) Now it's mostly a matter of expanding the terms and factoring to find solutions.
4) The only applicable solution I found was x=2
5) The upper limit with the x=2 substitution is indeed equal to 6.

Sorry for the short reply, I did it by hand and did the expansions and factoring by calculator since they were quite long. If you would like, I could write it out or post a picture of my solution if you did not understand my steps above.

Hope that helps,

Drood
Yes please, an image would help a lot. Thanks. Its just that its weird to have to solve it by guessing, but you seem to have some process there.
Drood
Drood is offline
#11
Feb26-12, 09:00 PM
P: 8
I hope you can read what I wrote. The site kills photo quality apparently and I didn't have enough time to copy it with my scanner.


Click image for larger version

Name:	IMAG0630.jpg
Views:	12
Size:	50.5 KB
ID:	44430

Hope that helps,

Drood
Psinter
Psinter is offline
#12
Feb26-12, 09:14 PM
Psinter's Avatar
P: 72
Quote Quote by Drood View Post
I hope you can read what I wrote. The site kills photo quality apparently and I didn't have enough time to copy it with my scanner.


Attachment 44430

Hope that helps,

Drood
Oh, I understand what you did there. Thank you very much. I feel dumb now because you make it look so easy and it took me so long.


Register to reply

Related Discussions
What is the distance moved up the plane (friction problem) Introductory Physics Homework 3
Particle is moved - Work-kinetic energy Introductory Physics Homework 1
Angle/distance (Theta) moved by a pendulum bob in t seconds. Classical Physics 4
Find time if boxes moved X distance Introductory Physics Homework 4
Distance moved as a function of time Advanced Physics Homework 1