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Distance traveled with linearly increasing kinetic energy

  • #1

Homework Statement


An object of mass 50 kg gains 20,000,000 joules every second. Devise formulae to find the distance covered at any given point in time, and the time necessary to cover a certain distance.

Homework Equations




The Attempt at a Solution


E = 20,000,000 * t
V = (E * 2/50)1/2
That's all I could come up with, couldn't get any further.
 

Answers and Replies

  • #2
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Homework Statement


An object of mass 50 kg gains 20,000,000 joules every second. Devise formulae to find the distance covered at any given point in time, and the time necessary to cover a certain distance.

Homework Equations




The Attempt at a Solution


E = 20,000,000 * t
V = (E * 2/50)1/2
That's all I could come up with, couldn't get any further.
Have you tried using calculus? We know that:
[tex] v = \frac{ds}{dt} [/tex]

Assuming the motion is in a straight line, displacement would be a suitable variable for distance. You could then integrate that to find an expression for s(t).

Is the question looking for two equations, because it seems that s(t) would satisfy both components of the question?
 
  • #3
PeroK
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@FutureOption do you know any calculus?

As suggested above, you will need calculus to solve this.
 
  • #4
No, haven't learned calc yet. I'll try looking things up and giving it a shot, though.
 
  • #5
PeroK
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No, haven't learned calc yet. I'll try looking things up and giving it a shot, though.
It's definitely a post-calculus problem.

PS I would solve the problem for mass ##m## (kg) and ##k## (J/s). Then plug in the specific numbers at the end. The calculus should be clearer with variables.
 
Last edited:
  • #6
Alright, it's definitely inelegant but I think I have something.
(20,000,000 * 2/50)1/2 = 894.427191

##\int_0^ x 894.427191 \cdot \sqrt{t} \, dt##

I believe that should give me the distance covered in a certain amount of time, am I on the right track?
 
  • #7
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Alright, it's definitely inelegant but I think I have something.
(20,000,000 * 2/50)1/2 = 894.427191

##\int_0^ x 894.427191 \cdot \sqrt{t} \, dt##

I believe that should give me the distance covered in a certain amount of time, am I on the right track?
Your limits should be 0 to t, not x. The reason is that we are integrating with respect to time (I am assuming the x was for distance/ displacement- if it was for time, then you are correct). We are looking for how much distance has been covered between the lower limit and the upper time limit, hence the start time (t=0) to a general time (t=t). Other than that, that would lead to the correct solution.

Sorry if I rambled on a bit, but given that you said that you were unfamiliar with calculus, I just wanted to try and explain the rationale behind the working. I hope that will help. Please do ask if it doesn't make sense.
 
  • #8
It's for time, I was told that the integration bounds shouldn't rely on the variable being integrated which is why I used x.
 
  • #9
PeroK
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Alright, it's definitely inelegant but I think I have something.
(20,000,000 * 2/50)1/2 = 894.427191

##\int_0^ x 894.427191 \cdot \sqrt{t} \, dt##

I believe that should give me the distance covered in a certain amount of time, am I on the right track?
See my comment above about using variables.

You still have some work to do in any case.

That is not an equation. You need an integral equation.
 
  • #10
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It's for time, I was told that the integration bounds shouldn't rely on the variable being integrated which is why I used x.
That is fair enough. It just prevents any careless errors by setting equal to x, rather than t. Both are perfectly acceptable though.
 
  • #11
PeroK
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Your limits should be 0 to t, not x. The reason is that we are integrating with respect to time (I am assuming the x was for distance/ displacement- if it was for time, then you are correct). We are looking for how much distance has been covered between the lower limit and the upper time limit, hence the start time (t=0) to a general time (t=t). Other than that, that would lead to the correct solution.

Sorry if I rambled on a bit, but given that you said that you were unfamiliar with calculus, I just wanted to try and explain the rationale behind the working. I hope that will help. Please do ask if it doesn't make sense.
This is not correct. One side will end up as an integral in ##x## or another dummy variable, and one side an integral in ##t##.
 
  • #12
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This is not correct. One side will end up as an integral in ##x## or another dummy variable, and one side an integral in ##t##.
What was incorrect about it?

I do not mean this in a rude way, just out of curiosity.
 
  • #13
Ray Vickson
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Alright, it's definitely inelegant but I think I have something.
(20,000,000 * 2/50)1/2 = 894.427191

##\int_0^ x 894.427191 \cdot \sqrt{t} \, dt##

I believe that should give me the distance covered in a certain amount of time, am I on the right track?
Please, please, please avoid using the symbol ##x## to stand for a time in a problem involving both time and distance; use some other t-related symbol, such as ##T## or ##t_1## of ##\tau## as your upper limit in the integration. Better still, use ##t## as the upper limit and use some other symbol for the "dummy" variable of integration, such as ##t'## or ##s## or ##\tau## (my favorite). Also, you can save space make things more clear by using a symbol such as ##c## instead of ##894.427191##. Carry out the calculation all the way to the very end, and only then plug in the numerical value of ##c##. So, if I were doing it I would write the distance ##x## at time ##t## as
$$x = \int_0^t c \sqrt{\tau} \, d\tau,$$
where ##c = 894.427191.##

Note: if you really DID mean to use distance ##x## as your upper integration limit, that would be 100% wrong.
 
  • #14
I appreciate the advice. I ended up interpreting PeroK's post as the opposite of what he meant which is why I used the numeric value.
 
  • #15
PeroK
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What was incorrect about it?

I do not mean this in a rude way, just out of curiosity.
There should be an integral equation somewhere. Not simply an integral in ##t##.
 
  • #16
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There should be an integral equation somewhere. Not simply an integral in ##t##.
Of course, but I assumed that they had simplified the integral
[tex] \int_{0}^{s} ds [/tex] to s and he was only posting about the RHS of the eqn
 
  • #17
So is it supposed to look something like this?
##s(t)=\int_0^t c * \sqrt{T} \, dT##

And about the second part of the question, presumably I need another equation for that. How do I go about getting it?
 
  • #18
PeroK
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So is it supposed to look something like this?
##s(t)=\int_0^t c * \sqrt{T} \, dT##

And about the second part of the question, presumably I need another equation for that. How do I go about getting it?
Yes. Now you have to integrate.
 
  • #19
##s(t)=c*\frac {t^{3/2}} {3/2}##
I believe that's it.
 
  • #20
Alright, looks like that is indeed the answer. Getting the second equation from there is pretty easy. Thanks, guys.
 

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