A new take on a relative velocity problem

[Differentiate it]

I can't explain it more until you start using $T$ when you mean the time of impact for a given $u$

Uhhh, just use T and explain..?

 

[PeroK]

Uhhh, just use T and explain..?

I need to see an attempt on your part to understand the difference between the continuous time coordinate $t$ and the time of impact $T$.

 

[Differentiate it]

I need to see an attempt on your part to understand the difference between the continuous time coordinate $t$ and the time of impact $T$.

could you at least answer the question I made a posts ago? I know its probably wrong, but just want to know if its correct, also I don't understand how u and t both are fixed when we fix v

 

[Shreya]

I need to see an attempt on your part to understand the difference between the continuous time coordinate t and the time of impact T

I think he just wants you to re-ask your question in post #49 by replacing $t$ with $T$ wherever applicable. Example, You have - "$t$ are maximum" $\rightarrow$ $T$ are maximum.
And In post #53, you said "$u$ and $t$" are fixed, while you should have said $u$ and $T$ are fixed. (Cause any choice of $u$ or $v$ doesn't fix the independent $t$ but it does fix $T$)
It may not seem like a big difference - but it is. $t$ is the independent continuous variable as @PeroK mentioned while $T$ is the impact time.
Moreover you are minimizing $u(T)$ not $u(t)$ & the graphs given were of $u(T)$. Also note that $u$ doesn't depend on $t$ but It does on $T$
It's my fault - because I used the wrong notation first and caused this confusion .

 

[Orodruin]

at least 90km/h but I don't know something doesn't look 100% rigorous with this, I found the approach described in post #2 more rigorous than the approach of post #4.

However post #4 is what we call a "clever shortcut".

The approach in post #4 is perfectly rigorous. Changing frames is not fishy in any way. The speed vs height follows from the SUVAT equations with final speed set to zero at the limiting case.

 

[Delta2]

Changing frames is not fishy in any way

Yes ok, I guess it is just me. Never feel comfortable when we use multiple reference frame in a single problem. That's also one reason I can't study relativity (the other main reason , not knowing tensor calculus lol).

 

[Orodruin]

Yes ok, I guess it is just me. Never feel comfortable when we use multiple reference frame in a single problem. That's also one reason I can't study relativity

This piques my interest. Where is the limit? Are you fine with using different Cartesian coordinate systems in geometry? What about using spherical coordinates?

the other main reason , not knowing tensor calculus lol

I know a good book …

 

[Delta2]

This piques my interest. Where is the limit? Are you fine with using different Cartesian coordinate systems in geometry? What about using spherical coordinates?I know a good book …

I am getting confused even if we going to switch between cartesian and spherical coordinates for the same physical problem.

A book on tensor calculus or on "relativity without tensor calculus"?

 

[Orodruin]

A book on tensor calculus or on "relativity without tensor calculus"?

A good book for learning tensor calculus ...

Although to be fair, most of introductory special relativity is perfectly accessible with just vector analysis.

 

[Differentiate it]

I think he just wants you to re-ask your question in post #49 by replacing $t$ with $T$ wherever applicable. Example, You have - "$t$ are maximum" $\rightarrow$ $T$ are maximum.
And In post #53, you said "$u$ and $t$" are fixed, while you should have said $u$ and $T$ are fixed. (Cause any choice of $u$ or $v$ doesn't fix the independent $t$ but it does fix $T$)
It may not seem like a big difference - but it is. $t$ is the independent continuous variable as @PeroK mentioned while $T$ is the impact time.
Moreover you are minimizing $u(T)$ not $u(t)$ & the graphs given were of $u(T)$. Also note that $u$ doesn't depend on $t$ but It does on $T$
It's my fault - because I used the wrong notation first and caused this confusion .

Yes, that's what I meant

 

[Differentiate it]

So, is this correct?:
We substitute in u - gt for v so that the function still has a minima?
If yes, could you tell which kind of functions can't have a maxima/minima ?
Also, could you answer the question made in #49 ?

 

[PeroK]

So, is this correct?:
We substitute in u - gt for v so that the function still has a minima?

What function has a minimum?

Also, could you answer the question made in #49 ?

$u$ is not a function of $t$. $u$ is the initial velocity. The velocity $v(t)$ at any time $t$ is a function of $t$ given by: $v(t) = u - gt$, where $u$ and $g$ are constants.

If you differentiate this you get the acceleration: $a(t) = v'(t) = -g$.

 

[Differentiate it]

So, is this correct?:
We substitute in u - gt for v so that the function still has a minima?
If yes, could you tell which kind of functions can't have a maxima/minima

What function has a minimum?

$u$ is not a function of $t$. $u$ is the initial velocity. The velocity $v(t)$ at any time $t$ is a function of $t$ given by: $v(t) = u - gt$, where $u$ and $g$ are constants.

If you differentiate this you get the acceleration: $a(t) = v'(t) = -g$.

Like, a linear function has no minimum( like u = v - 10t) but substituting √(u^2 -20(25t+125)) for v in, and writing as a function of t, gives -125/t - 25 - 5t, which does have a minimum. So, is the point of substituting u - gt to ensure that the function has a minimum? If yes, could you please add what kinds of functions have a minimum?

 

[jbriggs444]

Yes ok, I guess it is just me. Never feel comfortable when we use multiple reference frame in a single problem. That's also one reason I can't study relativity (the other main reason , not knowing tensor calculus lol).

I feel oppositely. I will change a frame of reference in preference to doing a bit of algebra, even though it amounts to the same thing. It's just more intuitive.

 

[PeroK]

So, is this correct?:
We substitute in u - gt for v so that the function still has a minima?
If yes, could you tell which kind of functions can't have a maxima/minima

Like, a linear function has no minimum( like u = v - 10t) but substituting √(u^2 -20(25t+125)) for v in, and writing as a function of t, gives -125/t - 25 - 5t, which does have a minimum. So, is the point of substituting u - gt to ensure that the function has a minimum? If yes, could you please add what kinds of functions have a minimum?

I don't understand your questions and I'm not sure I can be of any help to you. Perhaps someone else can try to understand what you are asking.

 

[Differentiate it]

I don't understand your questions and I'm not sure I can be of any help to you. Perhaps someone else can try to understand what you are asking.

Uh, so, I'll use a different example.
If I substituted in 25 m/s for v, in v^2 = u^2 + 2as, where s = 25t + 125, I get graph #1. But instead if I were to substitute in u-gt for v, I'd get graph #2(that function has a minima).
And my question was, do we substitute u-gt for v, to get a function that has a minima?

 

[PeroK]

If I substituted in 25 m/s for v, in v^2 = u^2 + 2as, where s = 25t + 125, I get graph #1.

No. You get a point. There is no function there. No free variable.

 

[Differentiate it]

No. You get a point. There is no function there. No free variable.

Isn't a function literally right there? That's not a point... anyways that's besides the point(get it?)... Is the answer to my question a yes or no?(although still how does that function become a point...?)

 

[PeroK]

(although still how does that function become a point...?)

$y = x^2$ is a function. $(1, 4)$ is a point that lies on the graph of that function. If you specify $y = 1$ or $x = 4$, then you are talking about a point and not a function.

You can differentiate a function: $y' = 2x$, but you can't differentiate a point. You can, however, evaluate the derivative at a point $y'(1) = 2$.

 

[Differentiate it]

$y = x^2$ is a function. $(1, 4)$ is a point that lies on the graph of that function. If you specify $y = 1$ or $x = 4$, then you are talking about a point and not a function.

You can differentiate a function: $y' = 2x$, but you can't differentiate a point. You can, however, evaluate the derivative at a point $y'(1) = 2$.

Well even if i specify v there's u = f(t), which I can still graph, it's a function!

 

[PeroK]

Well even if i specify v there's u = f(t), which I can still graph, it's a function!

$u$ is not a function of $t$. $u$ is the (constant) initial value at $t =0$.

 

[Differentiate it]

$u$ is not a function of $t$. $u$ is the (constant) initial value at $t =0$.

How? Also pls answer the question in post, #66, it'll be really helpful

 

[Shreya]

Well even if i specify v there's u = f(t), which I can still graph, it's a function!

I think you meant $u=f(T)$ not $f(t)$. Remember,there's a difference between $t$ and $T$.
Remember that the distance the bullet travels is given. On specifying $v$ (along with $s$), $u$ is determined ie graph collapses. You still have to solve for $u$ using simultaneous equations (it is just unknown but fixed). The graph doesn't take into account the $s$ therefore shows a function for $u(T)$. But you have to take $s$ into account.
@PeroK Please correct me if I am wrong.

 

[PeroK]

How?

By definition. I'm using the notation:
$$v(t) = u + at$$where $v(t)$ is the time dependent velocity and $u = v(0)$ is the initial velocity.

Note that although $v$ is a continuous function of $t$, we also tend to use $v$ for the "final" velocity. There is, therefore, a notational asymmtery between $u$, which is assumed to be fixed, and $v$, which is assumed to be a variable end-point.

 

[PeroK]

PS some people use $v_i$ and $v_f$ for initial and final velocities. But, this does not make as much difference as you might hope. We still have:
$$v_f = v_i + at$$And $v_f$ still manages to be a (variable) function of $t$!

 

[Differentiate it]

By definition. I'm using the notation:
$$v(t) = u + at$$where $v(t)$ is the time dependent velocity and $u = v(0)$ is the initial velocity.

Note that although $v$ is a continuous function of $t$, we also tend to use $v$ for the "final" velocity. There is, therefore, a notational asymmtery between $u$, which is assumed to be fixed, and $v$, which is assumed to be a variable end-point.

I haven't been keeping track or t and T, sorry. But at this point, just answer my question in post #66, and if the answer is no, please explain why. I'll be very thankful

 

[Herman Trivilino]

Ok I see now, basically the *intuitive* claim here is that the minimum speed is such that when the bullet reaches the balloon it has the same speed as the balloon. How do we justify this claim more formally?

That's the minimum speed needed to touch the balloon because with any speed less than that it wouldn't touch the balloon.

 

[Delta2]

That's the minimum speed needed to touch the balloon because with any speed less than that it wouldn't touch the balloon.

Hm ok but the approach described at post #2 solves the problem without the need to state this claim.

 

[Herman Trivilino]

Hm ok but the approach described at post #2 solves the problem without the need to state this claim.

It's assumed.

 

[Differentiate it]

@PeroK
My final question is the one in #66. If the answer to that is no, please explain why. That's the final question, I won't bug anyone after that

 

[Delta2]

It's assumed.

No it isn't assumed. You just say there $$h=25t+125$$ (height of balloon), $$h'=v_0t-0.5gt^2$$ (height of bullet) form the quadratic equation (with respect to t) $$h=h'$$ and the minimum requirement for $v_0$ pops in a "wonderous " way from the requirement that the discriminant $$(v_0-25)^2-4\cdot 0.5\cdot g \cdot 125\geq 0$$of the quadratic is greater or equal to zero.
No change of frames, no additional statements, almost pure math. That's why I found it more rigorous.

 

[PeroK]

I haven't been keeping track or t and T, sorry.

That's the key. Without understanding that, you are just asking fairly irrelevant questions about minimising functions in general.

The substitution $v = u - gt$ didn't change the nature of the equations, but replaced the variable $v$ by $u$ and $t$, because we wanted to solve for the impact time in terms of $u$. Having both $v$ and $u$ in the equation was not what we wanted.

The solution to that equation is $t = T$, which leads to a function for $u$ and $T$, which we can minimise. Note that $u$ is a function of $T$ but not of $t$.

 

[Differentiate it]

That's the key. Without understanding that, you are just asking fairly irrelevant questions about minimising functions in general.

The substitution $v = u - gt$ didn't change the nature of the equations, but replaced the variable $v$ by $u$ and $t$, because we wanted to solve for the impact time in terms of $u$. Having both $v$ and $u$ in the equation was not what we wanted.

The solution to that equation is $t = T$, which leads to a function for $u$ and $T$, which we can minimise. Note that $u$ is a function of $T$ but not of $t$.

I'm sorry, but please just answer my question from post #66. That's it.

 

[PeroK]

I'm sorry, but please just answer my question from post #66. That's it.

I don't understand your question. It makes no sense to me. I've explained why I used $v = u - gt$. We are not minimising SUVAT equations. How many times do you have to be told that?

 

[Differentiate it]

I don't understand your question. It makes no sense to me. I've explained why I used $v = u - gt$. We are not minimising SUVAT equations. How many times do you have to be told that?

Some functions, like linear ones, like u = 10t + 25, don't have a minima.(i got this from v = u +at where i fixed the value of v to 25 m/s) Other functions like 125/t + 25 + 5t do have a minima. So, instead of fixing v to a certain value, if we substitute v = sqrt(u^2 + 2as) for v in v = u +at where s = 25t + 125, and rearrange for initial velocity u, we get u = 125/t + 25 + 5t, which does have a minima. So, my final question was, do we substitute in something different for v(in this case it was sqrt(u^2 + 2as)) instead of a fixed value like 25 to ensure that the function 𝘩𝘢𝘴 𝘢 𝘮𝘪𝘯𝘪𝘮𝘢. I don't think this has anything to do with t or T, which I didn't keep track of, I'm sorry, but please, just answer this question

 

[Herman Trivilino]

So, my final question was, do we substitute in something different for v(in this case it was sqrt(u^2 + 2as)) instead of a fixed value like 25 to ensure that the function 𝘩𝘢𝘴 𝘢 𝘮𝘪𝘯𝘪𝘮𝘢.

It looks like that's what you did, so yes. But I would use the word "create" rather than "ensure".

 

[Differentiate it]

It looks like that's what you did, so yes. But I would use the word "create" rather than "ensure".

Thanks! And, sorry for my poor wording

 

[Differentiate it]

Thanks! And, sorry for my poor wording

I still don't understand how i would find such a function (if I knew why some functions didn't have a minima(due to physical reasons), i would know what type, scratch that, which equation to use)
@PeroK 's explanation made absolutely no sense to me. If anyone could tell me the physical reason why, I'll be very thankful

 

[neilparker62]

See following graph(s) for vertical displacement (vs time) of bullet and balloon. We need to set $y_a-y_b = 0$ and further note that for the bullet to just touch the balloon, the (quadratic) equation will have real and equal roots. ie $\Delta=b^2-4ac=0$.

https://www.desmos.com/calculator/0iwexj13pz