Find the distance a particle travels

[figuringphysics]

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

 

[nasu]

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.

 

[WWGD]

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.

 

[PeroK]

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 $m/s^3$.

 

[Orodruin]

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…

 

[PeroK]

This has dimensions of … length…

What a jerk I've been!

 

[Orodruin]

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 …

 

[figuringphysics]

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:

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 :)

 

[figuringphysics]

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?

 

[figuringphysics]

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!

 

[PeroK]

I see, will remember to include units in the future.

It's more the people who set these questions teaching you bad habits.

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.

 

[PeroK]

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.

 

[MatinSAR]

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.

 

[WWGD]

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.

 

[Herman Trivilino]

how it is possible to "reset" the time like this for the second equation?

Instead of writing $v=3e^{-t}$ you would write $v=3e^{-(t-1)}$.

 

[jbriggs444]

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.