Vertical displacement with time of a projectile

[pbuk]

That contradicts your second assertion that the object has no vertical velocity for $t > 0$.

Where did I assert that? I asserted (or rather I repeated the statement in the question) that the object has no vertical velocity for t = 0.

 

[pbuk]

You cannot have it both ways. If being at rest at $t=0$ with a discontinuous change at $t>0$ is allowed for the horizontal velocity then surely the same can be true for the vertical velocity.

That would be the case if being at rest at $t = 0$ were the only condition, however there is another condition that vertical acceleration is constant, ruling out an initial discontinuity.

 

[PeroK]

Where did I assert that? I asserted (or rather I repeated the statement in the question) that the object has no vertical velocity for t = 0.

In your argument to rule out graph B?

Anyway, that fact that three senior HH are criticising this problem is enough in itself.

That would be the case if being at rest at $t = 0$ were the only condition, however there is another condition that vertical acceleration is constant, ruling out an initial discontinuity.

This is my objection. That's an arbirrary interpretation of kinematic data. A linear graph for horizontal displacement does not represent an object initially at rest. A linear graph represents an object with an initial horizontal component of velocity.

 

[jbriggs444]

That would be the case if being at rest at $t = 0$ were the only condition, however there is another condition that vertical acceleration is constant, ruling out an initial discontinuity.

Technically, that is not the case.

If the object is "at rest" at $t=0$ this means that we are contemplating times before the start of the problem. If we were only contemplating $t \ge 0$ then there would be no sense in which the object is at rest. Its horizontal velocity is constant at $t>0$. Its [one-sided] horizontal velocity at $t=0$ is the same.

But you say that the object is initially at rest. So we must be contemplating negative times.

Velocity is the first derivative of displacement. It cannot change discontinuously from one constant value to another. The mean value theorem forbids it. So horizontal velocity is undefined at $t=0$.

The same argument applied to vertical acceleration. It is a derivative. It cannot change discontinuously from one constant value to another. The mean value theorem forbids it. So vertical acceleration is undefined at $t=0$.

So there is no initial discontinuity. Only a point at which both velocity and acceleration are undefined. Which makes the stipulation that the object begins at rest irrelevant.

 

[pbuk]

In your argument to rule out graph B?

Are you referring to this:

But you can't have a .... non-zero initial vertical component, and so B is not valid.

Here I talk about the initial velocity i.e. at $t = 0$, not when $t > 0$.

Anyway, that fact that three senior HH are criticising this problem is enough in itself.

Not all HH in this thread share your view that the problem is "a mess".

This is my objection. That's an arbirrary interpretation of kinematic data. A linear graph for horizontal displacement does not represent an object initially at rest. A linear graph represents an object with an initial horizontal component of velocity.

But there is no linear graph!

 

[jbriggs444]

Are you referring to this:

Here I talk about the initial velocity i.e. at $t = 0$, not when $t > 0$.

Velocity cannot make a step change without going undefined. If you claim a step change at $t=0$, you must abandon a claim about velocity at $t=0$.

 

[Herman Trivilino]

It is not clear what is meant by "moves from rest". That conflicts with "constant horizontal velocity" unless that is zero.

Okay. You've convinced me that I'm wrong. Moves from rest means starts moving from a state of rest. That's not possible if the horizontal velocity is constant.

The question is indeed incorrectly written. It could be fixed by stating that the object has zero horizontal acceleration. Then the question would probe what appears to have been intended.

 

[pbuk]

Thank you @jbriggs444 for your considered analysis. As it happens I agree with you that

the wording is bad.

, however that does not mean that option B is a correct answer (and I don't think you are arguing that it is?)

But I do not agree with the rest of your analysis.

If the object is "at rest" at $t=0$ this means that we are contemplating times before the start of the problem. If we were only contemplating $t \ge 0$ then there would be no sense in which the object is at rest.

I don't agree with this. For me "at rest" at $t=0$ means only that $t = 0 \implies v = 0$.

Velocity is the first derivative of displacement.

Where displacement is differentiable that is true, however I do not agree that (in an ideal model) that displacement must be differentiable everywhere: for instance our ideal model of a collision prevents that.

It cannot change discontinuously from one constant value to another. The mean value theorem forbids it.

The mean value theorem requires a differentiable (and therefore continuous) function so you can't invoke it to prove differentiability.

So horizontal velocity is undefined at $t=0$.

It is defined as zero in the question.

The same argument applied to vertical acceleration. It is a derivative. It cannot change discontinuously from one constant value to another.

What happens when we cut a string suspending an object?

 

[PeroK]

Not all HH in this thread share your view that the problem is "a mess".

That doesn't matter. If I wrote a question and gave it to five judges, and two thought it made sense and three thought it didn't, then the question is problematic. It would be wrong to pass that question on the grounds that two out of five judges understood it. At the very least, that suggests the question would make sense to a minority of students.

If it was only me that found the question unsatisfactory, then you could argue a case for it. But, three out of five of us think it has potential to confuse. One problem is that if the student is subsequently faced with a linear graph of displacement against time, is that a graph of:

a) Constant velocity.

b) Constant velocity from a state of rest.

c) Both of the above.

I don't like the question. I think it should be revised.

 

[pbuk]

But, three out of five of us think it has potential to confuse.

Oh let's be clear, I also think it has potential to confuse, and if I was writing a text book (which is where it seems to have come from) I would not have worded it that way. However that does not mean that answer B should be allowed as a valid answer to the question as it is worded, which is what you and @haruspex are arguing.

I don't like the question. I think it should be revised.

Do you think that any discussion here will achieve that?

 

[PeroK]

However that does not mean that answer B should be allowed as a valid answer to the question as it is worded, which is what you and @haruspex are arguing.

I'm not arguing that. Graph D is the only possibility. But, I don't like the question because there is no motion in the horizontal direction. It feels like something that has been cobbled together with too little thought.

 

[pbuk]

Velocity cannot make a step change without going undefined. If you claim a step change at $t=0$, you must abandon a claim about velocity at $t=0$.

That is true: because the question implies that there is (possibly a) step change in the horizontal component of velocity at $t = 0$ (from 0 to some, possibly non-zero, constant), the horizontal component is undefined at $t = 0$.

However this is irrelevant to the vertical component of velocity where there is no step change - the question tells us that this is 0 at $t = 0$ and increases with some (presumably finite) constant acceleration.

 

[pbuk]

I'm not arguing that. Graph D is the only possibility.

Ah, when I read posts #12 and #13 I got the impression that you were also arguing for graph B. This is the only reason I posted on this thread - to try to prevent a student from believing that graph B should be allowed as a correct answer.

 

[erobz]

Everyone likes a good sparring now and then...keeps the skills sharp. Intellectual ones are less exhilarating than the physical, but someone still gets their brain beat out!

 

[jbriggs444]

Thank you @jbriggs444 for your considered analysis. As it happens I agree with you that

, however that does not mean that option B is a correct answer (and I don't think you are arguing that it is?)

But I do not agree with the rest of your analysis.


I don't agree with this. For me "at rest" at $t=0$ means only that $t = 0 \implies v = 0$.

The mathematical definition of the first derivative is in conflict with this.$$\lim_{h \to 0} \frac{f(a+h) - f(a)}{h}$$With the understanding that $a$ and $a+h$ are constrained to fall within the domain of $f$.

If the two one-sided derivatives disagree then the two sided derivative does not exist.

Where displacement is differentiable that is true, however I do not agree that (in an ideal model) that displacement must be differentiable everywhere: for instance our ideal model of a collision prevents that.

Right. If velocity changes around zero then the position function is not differentiable at zero. So $v(t)$ is undefined at $t=0$. Not zero. Undefined.

The mean value theorem requires a differentiable (and therefore continuous) function so you can't invoke it to prove differentiability.

I am invoking it to prove the lack of differentiability.

The mean value theorem requires not just differentiability. It also requires that the function be defined everywhere within a closed interval and that it be differentiable everywhere within the open interval. I am invoking it to prove that the derivative is not defined throughout an open interval that encompasses both positive and negative times. In particular, that it is undefined at $t=0$.

Let me take you through it. We will be dealing here with horizontal velocity.

I will denote position as $s(t)$ and its first derivative as $v(t)$.

I will take "at rest" to mean that $s(t) = 0$ throughout some closed interval ending at $t=0$. This implies that $v(t)=0$ throughout the open interval ending at $t=0$. I make no inference about $v(0)$.

Let us denote the start of this interval as $a$.

I will take "constant velocity" to mean that $v(t) = k > 0$ throughout some open interval beginning at $t=0$. I make no inference about $v(0)$.

Let us denote the end of this interval as $b$.

I think you will grant me that the position function is continuous and well defined over the closed interval $[a,b]$ and that $s(b) > s(a)$. I think that you will also grant me that the average velocity over the interval is less than $k$, the constant velocity for $t > 0$.

That is:$$\frac{s(b) - s(a)}{b-a} < k$$At this point, we have almost everything we need to invoke the mean value theorem. We have a function that is defined and continuous on a closed interval. It is differentiable on the open interval except, possibly at $t=0$

If $v(t)$ is defined for $t=0$ we will have everything we need for the mean value theorem. So let us suppose (for purposes of a proof by contradiction) that $v(t)$ is defined for $t=0$.

Then it follows from the mean value theorem that $v(t) = \frac{s(b) - s(a)}{b-a}$ for some $t$ in the open interval $(a,b)$

But we've already said that $v(t) = 0$ for $t<0$ in this interval. And we've already said that $v(t) = k$ for $t>0$ in this interval.

If $v(0)= 0$ the mean value theorem is falsified.
If $v(0) = k$ the mean value theorem is falsified.

There is a loophole if $v(0) = \frac{s(b) - s(a)}{b-a}$. This is tedious to address and is obviously silly. So let us ignore it.

Contradiction. So $v(0)$ is undefined.

It is defined as zero in the question.

No. It is not. The question says "at rest". The implication is that the derivative is zero at $t < 0$. The implication does not hold for $t=0$.

What happens when we cut a string suspending an object?

The acceleration changes suddenly. It is zero before the cut. It is non-zero and constant after the cut. The acceleration at the moment the string is cut is undefined. Two one-sided second derivatives of position exist and are different.

The velocity does not change suddenly. It is zero both before and after the cut. (To be picky, after the cut it is non-zero and increasing with a limit of zero as the time of the cut is approached). The velocity at the time of the cut is well defined and is zero.

The position does not change suddenly.

 

[Gargi]

Homework Statement: image below
Relevant Equations: suvat equations

View attachment 342725
my answer was D, which is correct considering the motion downwards eg, object thrown from a cliff
but what if an object is moving up?like a football being kicked?
i cant understand how this graph applies to that...

To kick a football, we will require an initial velocity in the vertical direction as well which is not the case here. In this case, the vertical displacement would occur according to y = 1/2at^2 which is an upward parabolic curve

 

[Herman Trivilino]

I don't agree with this. For me "at rest" at $t=0$ means only that $t = 0 \implies v = 0$.

Typically, at rest means the speed is zero for some nonzero length of time.

What happens when we cut a string suspending an object?

Typically we model that as an event. Events have a time duration of zero. So, before the string is cut the object is at rest. When the string is cut the speed is zero at that instant, but not for any duration of time. As soon as the string is cut the speed begins to increase.

Another example is a ball tossed vertically upward. At the instant it reaches its apex, the speed is zero, but that is typically not considered a state of rest, because it lasts for no length of time at all.

Of course our language can be used differently, but that's a matter of semantics. I'm just referring to what I've found to be overwhelmingly typical in the college-level introductory physics textbooks.

 

[berkeman]

Thread closed for Moderation...

 

[berkeman]

Everyone likes a good sparring now and then...keeps the skills sharp.

Except that it doesn't necessarily help the OP. I think I'll keep this thread closed since the OP seems not to be interested anymore. If you folks want to start a separate thread to discuss issues with the question, that would probably be the best way to handle it. Thanks.