Find the distance a particle travels

  • #1
figuringphysics
4
0
Homework Statement
A particle travels at velocity v(t)=3t^2 for 1s and then at v(t) = 3e^-t forever. How far does it travel?
Relevant Equations
x(t) = ∫ v(t) dt
My answer is d = (e+3)/e

x(t) = ∫01 3t2 dt (0 ≤ t ≤1)
= t3 |01
= 13 - 03
= 1m

x(t) = ∫1 3e-t dt (t > 1)
= -3e-t |1
= lim(t->∞)[-3e-t] - [-3e-1]
= 0 + 3e-1
= 3/e m

Therefore total distance = 1m + 3/e m = (e+3)/e m

However, the textbook answer gives 4m. I can see how this is possible if we integrate the second equation from 0 to infinity:

x(t) = ∫0 v(t) dt (1 < t)
= -3e-t |0
= 3 m

But I don't think this is correct. Am I missing something in the question? Thanks
 
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  • #2
There is some problem in your problem statement. What is the velocity at t=1? The two formulas, as given, do not match at t=1. Unless the time in the second one is measured from t=1.
 
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  • #3
It may be the case, through reverse engineering, that you're expected to start counting from t= 0 after that one second. You get the right answer. Otherwise, as @nasu points out, you have a discontinuity at t=1 second, where your velocity is both 1, according to the first formula, and 3/e according to the second.
 
  • #4
figuringphysics said:
Homework Statement: A particle travels at velocity v(t)=3t^2 for 1s and then at v(t) = 3e^-t forever. How far does it travel?
Note also that these formulas are dimensonally inconsistent. A velocity cannot equal a time squared. The first one can be fixed by identifying the ##3## as an acceleration:
$$v(t) = (3 m/s^2)t^2$$The second one could be written as:
$$v(t) = (3m/s)e^{-(t -1s)/(1s)}$$PS as @Orodruin points out, the ##3## is actally a jerk and has units of ##m/s^3##.
 
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  • #5
PeroK said:
The first one can be fixed by identifying the ##3## as an acceleration:
$$v(t) = (3 m/s^2)t^2$$
This has dimensions of … length… 😏
 
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  • #6
Orodruin said:
This has dimensions of … length… 😏
What a jerk I've been!
 
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  • #7
PeroK said:
What a jerk I've been!
I cannot decide what is worse: The original transgression or that joke … or the fact that it made me laugh …
 
  • #8
nasu said:
There is some problem in your problem statement. What is the velocity at t=1? The two formulas, as given, do not match at t=1. Unless the time in the second one is measured from t=1.
The velocity is 3 m/s according to the first equation, and 3/e m/s according to the second:
1710443474472.png


We would "solve" this discontinuity by adding a term representing the initial velocity (##v(x)=3e^{-t} + \frac{3e-3}{e}##), but then integrating gives: $$x(t)=-3e^{-t} + \frac{3e-3}{e}t + C_1$$ which doesn't approach a limit as ##t\rightarrow\infty## right? Thanks :)
 
  • #9
WWGD said:
It may be the case, through reverse engineering, that you're expected to start counting from t= 0 after that one second. You get the right answer. Otherwise, as @nasu points out, you have a discontinuity at t=1 second, where your velocity is both 1, according to the first formula, and 3/e according to the second.
Thank you. I'm just not sure how we know to integrate from t=0s for the second equation, how it is possible to "reset" the time like this for the second equation?
 
  • #10
PeroK said:
Note also that these formulas are dimensonally inconsistent. A velocity cannot equal a time squared. The first one can be fixed by identifying the ##3## as an acceleration:
$$v(t) = (3 m/s^2)t^2$$The second one could be written as:
$$v(t) = (3m/s)e^{-(t -1s)/(1s)}$$PS as @Orodruin points out, the ##3## is actally a jerk and has units of ##m/s^3##.
I see, will remember to include units in the future. The change to the second equation is essentially shifting it to the right by 1s to maintain continuity if I'm understanding correctly - is it valid to do this in this situation?

Related to my reply to @WWGD.

And following from this, would that make my original answer of ##\frac{e+3}{e}m## incorrect, due to the discontinuity? Thanks for answering all my questions!
 
  • #11
figuringphysics said:
I see, will remember to include units in the future.
It's more the people who set these questions teaching you bad habits.
figuringphysics said:
The change to the second equation is essentially shifting it to the right by 1s to maintain continuity if I'm understanding correctly - is it valid to do this in this situation?

And following from this, would that make my original answer of ##\frac{e+3}{e}m## incorrect, due to the discontinuity? Thanks for answering all my questions!
Your answer is not really wrong. But, once you see the textbook answer of ##4m##, you should be able to re-interpret the question. And the second equation intended you to reset the clock to ##t = 0##.

A poorly set question, if you ask me.
 
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  • #12
figuringphysics said:
Thank you. I'm just not sure how we know to integrate from t=0s for the second equation, how it is possible to "reset" the time like this for the second equation?
We know because it gives the right answer. It's unusual for a problem to assume you reset the clock between two phases of motion without indicating this explicitly.

It also didn't say that the first phase of motion started at ##t = 0s##. Although, that would be a fair assumption. It only said for ##1s##. It didn't say from ##t = 0s## to ##t = 1s##. That lack of attention to detail is partly why the question setter went wrong. They had an idea in mind that they failed to communicate in the problem statement.
 
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  • #13
figuringphysics said:
Homework Statement: A particle travels at velocity v(t)=3t^2 for 1s and then at v(t) = 3e^-t forever. How far does it travel?
Relevant Equations: x(t) = ∫ v(t) dt

But I don't think this is correct. Am I missing something in the question? Thanks
I can't find out how this particle's velocity has two values at t=1. As far as I know this is impossible.
Even for very very small changes in velocity, Time won't stop and it passes.
 
  • #14
MatinSAR said:
I can't find out how this particle's velocity has two values at t=1. As far as I know this is impossible.
Even for very very small changes in velocity, Time won't stop and it passes.
You'd need an insanely high acceleration if you just consider ##3^{-}## vs ##3^{+}##, when approaching left or right. But it would make sense for the second formular to be self-contained or " self-referential ", in that it starts at 0, sort of ignoring what's happened prior. But the author could have helped in clarifying.
 
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  • #15
figuringphysics said:
how it is possible to "reset" the time like this for the second equation?
Instead of writing ##v=3e^{-t}## you would write ##v=3e^{-(t-1)}##.
 
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  • #16
MatinSAR said:
I can't find out how this particle's velocity has two values at t=1. As far as I know this is impossible.
You are correct that it is impossible. A discontinuous change in first derivative like this would be a violation of the mean value theorem.
 
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