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

In summary: I computed the integral and applied the limits.2) I know that the integral with the limit substitutions = 1083) Now it's mostly a matter of expanding the terms and factoring to find solutions.4) The only applicable solution I found was x=25) The upper limit with the x=2 substitution is indeed equal to 6.
  • #1
Psinter
278
787

Homework Statement


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


Homework Equations


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


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. :grumpy:
 
Physics news on Phys.org
  • #2
What is the integral of x2 + 2x?
 
  • #3
alanlu said:
What is the integral of x2 + 2x?
Its [itex]\frac{1}{3}x^3+x^2[/itex]
 
  • #4
Psinter said:

Homework Statement


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


Homework Equations


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


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. :grumpy:

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
 
  • #5
Ray Vickson said:
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. :frown:

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?
 
Last edited:
  • #6
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.
 
  • #7
alanlu said:
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.
 
  • #8
Psinter said:
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
 
  • #9
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
 
  • #10
Ray Vickson said:
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.

Drood said:
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.
 
Last edited:
  • #11
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.


IMAG0630.jpg


Hope that helps,

Drood
 
  • #12
Drood said:
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.View 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.
 

1. What is the relationship between work and distance for a moving particle?

The work done on a particle is directly proportional to the distance it has moved. This means that as the work done on the particle increases, so does the distance it has moved.

2. How do you calculate the distance a particle has moved if the work done is known?

The distance a particle has moved can be calculated using the equation W = Fd, where W is the work done, F is the force applied, and d is the distance moved. Rearranging the equation, we get d = W/F.

3. Does the direction of the work done affect the distance a particle has moved?

Yes, the direction of the work done does affect the distance a particle has moved. If the work done is in the same direction as the motion, the distance moved will be greater. However, if the work done is in the opposite direction of the motion, the distance moved will be less.

4. Can the distance a particle has moved be negative?

No, the distance a particle has moved cannot be negative. Distance is a scalar quantity and is always positive. If the particle moves in the opposite direction of the work done, the distance can be represented as a negative value of displacement.

5. Is the distance a particle has moved the same as its displacement?

No, distance and displacement are two different quantities. Distance is the total length of the path traveled, while displacement is the shortest distance between the starting and ending points. In some cases, the distance and displacement may be the same, but in most cases, they will be different.

Similar threads

  • Calculus and Beyond Homework Help
Replies
2
Views
452
  • Introductory Physics Homework Help
Replies
25
Views
908
  • Calculus and Beyond Homework Help
Replies
2
Views
3K
  • Calculus and Beyond Homework Help
Replies
2
Views
826
  • Introductory Physics Homework Help
Replies
28
Views
338
  • Introductory Physics Homework Help
Replies
5
Views
503
  • Calculus and Beyond Homework Help
Replies
19
Views
2K
  • Calculus and Beyond Homework Help
Replies
5
Views
1K
  • Calculus and Beyond Homework Help
Replies
6
Views
2K
  • Calculus and Beyond Homework Help
Replies
14
Views
1K
Back
Top