Calculating velocity from the position versus time graph

[fmiren]

Homework Statement

a) Identify
the time or times ( ta , tb , tc , etc.) at which the instantaneous velocity is greatest. (b) At which times is it zero? (c) At which
times is it negative?

Relevant Equations

instantaneous velocity

My answers: a) At points d the instantaneous velocity should be greatest since slope of c-d is greatest I think
b) At point e and g instantaneous velocity is 0
c) at points b, c, and f instantaneous velocity is negative.

Could you please verify my answers?

 

[hmmm27]

Are you sure those are the answers you want to give ? For starters, the question clearly indicates that "point x" be written as "tx".

 

[fmiren]

Yes, I'm afraid. The only correction that I can make is to a) part: at tl instantaneous velocity should also be 0.
Where I'm wrong?

 

[etotheipi]

I don't know about you, but I'd say $t=t_c$ corresponds to a local minimum of position.

 

[fmiren]

Yes, I think at t_c it is also 0.

So my answers:
a) t_d
b) t_c, t_e, t_g, t_l
c) t_b, t_f

Is it correct now?

 

[jbriggs444]

So my answers:
a) t_d
b) t_c, t_e, t_g, t_l
c) t_b, t_f

Is it correct now?

I do not agree with the stated result for c).

 

[hmmm27]

I'm wondering why - in b) - you include $t_l$ and not $t_a$ (not saying it's wrong or right, just wanting some justification)

 

[fmiren]

I'm wondering why - in b) - you include $t_l$ and not $t_a$ (not saying it's wrong or right, just wanting some justification)

To be honest, I didn't pay enough attention to it. Now looking at again, I think, yes, t_a should be included.
But I don't understand why the answer for c) is wrong. Any hint please?

 

[haruspex]

But I don't understand why the answer for c) is wrong. Any hint please?

What is the slope at a? What is the slope at c?

 

[fmiren]

At t_a it is negative, at t_c it is zero.

 

[haruspex]

At t_a it is negative, at t_c it is zero.

So why do you not list a as a point where the velocity is negative?

 

[fmiren]

Yes, in one of comments above I included it.
a) t_d
b) t_c, t_e, t_g, t_l
c) t_a, t_b, t_f

Is it ok now?

 

[hmmm27]

Yes, in one of comments above I included it.

You sure about that ?

Is it ok now?

Not to get ahead of the game too much, but $t_a$ and $t_l$ are discontinuities. Depending on what course the problem is a part of, or what lesson is being taught, they may want you to treat them as special cases.

 

[jbriggs444]

Yes, in one of comments above I included it.
a) t_d
b) t_c, t_e, t_g, t_l
c) t_a, t_b, t_f

Is it ok now?

Looks good to me.

 

[fmiren]

You sure about that ?
Not to get ahead of the game too much, but $t_a$ and $t_l$ are discontinuities. Depending on what course the problem is a part of, or what lesson is being taught, they may want you to treat them as special cases.

It is from a open textbook "College Physics" - https://openstax.org/details/books/college-physics

I don't remember any mention of special cases in the text. Could you please explain what they are?

 

[etotheipi]

I don't remember any mention of special cases in the text. Could you please explain what they are?

If the function is discontinuous at that point, it's also not differentiable at that point.

 

[fmiren]

If the function is discontinuous at that point, it's also not differentiable at that point.

Thank you. How does it apply to this problem? Is the function discontinuous at c, e, g, and l points? And what it implies?

 

[hmmm27]

Thank you. How does it apply to this problem? Is the function discontinuous at c, e, g, and l points? And what it implies?

You're welcome. It would apply if the course was Introduction to Calculus : my apologies for introducing irrelevancies. The definition of "discontinuity" in context is the same as in common language discourse. @etotheipi answered that one in post #16.

 

[jbriggs444]

Thank you. How does it apply to this problem? Is the function discontinuous at c, e, g, and l points? And what it implies?

This has to do with picky little details about the definition of continuity. It has little to do with the problem posed in your original post.

Specifically, we are getting at the question of whether a function is continuous at its endpoints.

Intuitively, for a function to be continuous at a point, the graph of that function must not have a break at that point -- you should be able to draw the graph without lifting your pencil from the paper. You are also not allowed to have any perfectly vertical lines on the graph, but that's part of what it means to have a function in the first place.

When you get into a math course where continuity is more carefully defined you find a definition along the lines of:

A function f() is continuous at a point l iff

1. l is in the domain of f
2. for all epsilon > 0 there is a delta > 0 such that for all x in the domain of f such that | x - l | < delta, | f(x) - f(l) | < epsilon

Paraphrased, that says if you look in a neighborhood that is close to l, function values in that neighborhood will be close to f(l). No matter how close you need the function values to be, you can always find a neighborhood that is small enough to be sure that the function values in that neighborhood will all be that close.

The point here is that it does not matter for the purposes of continuity whether you are at the edge of the function's domain or not. x values that do not fall within the function's domain do not count against continuity.

Wiki points this out:

"It follows from this definition that a function f is automatically continuous at every isolated point of its domain. As a specific example, every real valued function on the set of integers is continuous."

 

[jbriggs444]

If the function is discontinuous at that point, it's also not differentiable at that point.

The function here is continuous throughout its domain. So this fact, while correct, is irrelevant.

What is relevant is the definition of a derivative.

"Let f be a real valued function defined in an open neighborhood of a real number a..."

At the endpoints of an interval you are not going to satisfy such a definition.

Whether one uses this fact to make a claim that "velocity" is undefined at the endpoints of the given function is a matter of taste. Personally, I am happy to accept the existence of a velocity based on a one-sided derivative at the endpoints even though the two-sided derivative is formally undefined.

 

[hmmm27]

Personally, I am happy to accept the existence of a velocity based on a one-sided derivative at the endpoints even though the two-sided derivative is formally undefined.

Yes, but you've been at it for decades ; the OP for hours, at most, and - as it turns out - it's not near-immediate to the course of study.