A new take on a relative velocity problem

[Shreya]

Because v=25 m/s implies the specific solution and determines u and t. These are then no longer continuous variables. If v=25m/s, then u=75 m/s and t=5 s and there is no function to differentiate.

Awesome! Does that mean that the equations we get in 3 & 4 doesn't describe the range of $u$s and the corresponding $t$s we can get, but only a specific situation.
Another Silly Question - then why do we still get a function (not a number)? And doesn't the lower values of u from 3 & 4 satisfy the conditions?

 

[Shreya]

I did not contribute much other than trying to ensure balloon popping! But thanks for including me in the "credits" list all the same.!

I didnt contribute much either, I think 95% of the thanks should go to Perok, Andrewkirk and Kuruman

For a student who is self learning physics with no conventional teacher & only access to internet, and who has spent more than 3 hrs on this problem myself with no progress, anybody who has contributed is godlike

 

[Differentiate it]

Because $v = 25 \ m/s$ implies the specific solution and determines $u$ and $t$. These are then no longer continuous variables. If $v = 25 m/s$, then $u = 75 \ m/s$ and $t = 5 \ s$ and there is no function to differentiate.

But u and t are both unknown. u can be any value and t can be any value too, right?

 

[Differentiate it]

That equation applies to the bullet regardless of what the balloon is doing.

Hm, but how does s = ut + (1/2)at^2 take the balloon's motion into account? s is the displacement of the bullet(displacement of the balloon from when the bullet is fired to when it hits the balloon is just 25t), u is the bullets initial velocity, and t is the time taken to hit the balloon

 

[Shreya]

Hm, but how does s = ut + (1/2)at^2 take the balloon's motion into account? s is the displacement of the bullet(displacement of the balloon from when the bullet is fired to when it hits the balloon is just 25t), u is the bullets initial velocity, and t is the time taken to hit the

$v=u+at$ just takes into account the fact that the bullet experiences deceleration and will have a final Velocity of $v$ later. It is too general and doesn't take into account the distance the bullet has to cover to pop the balloon (or anything regarding the situation), while $s=ut+\frac{1}{2}at²$ states that the bullet has to cover the required distance while experiencing deceleration.
Note that $v=u+at$ can be used for any object moving up against gravity and reaches the said $v$ - it doesn't have what we are looking for ie the bullet has to pop the balloon

 

[PeroK]

Awesome! Does that mean that the equations we get in 3 & 4 doesn't describe the range of $u$s and the corresponding $t$s we can get, but only a specific situation.
Another Silly Question - then why do we still get a function (not a number)? And doesn't the lower values of u from 3 & 4 satisfy the conditions?

Essentially you do get a number (really a quantity, as we have units involved). There's a difference between a fixed, as yet uncalculated quantity and a variable. To take a simple example.

We start with the simple equation relating displacement, constant velocity and time:
$$s = vt$$In general, all of these quantities may be variables. Note that there is a subtle difference between $v$, which is an unknown, and $s$ and $t$ which are continuous variables.

In a single experiment, $v$ would be some fixed quantity throughout the experiment, while $t$ and $s$ vary.

If we consider varying $v$, then we are considering many different experiments. This is the technique you tried to use for this problem. This is a valid approach, but you must be aware of what you are doing.

If we want to hit a target at a distance of $10 \ m$ then the speed and time are related by:
$$vT = 10 \ $$Which gives $v$ as a function of $T$ and vice versa. We can plot a graph of $v$ against $T$. Note that I've been careful to use $T$ here to indicate that we have a fixed but unknown time $T$. In any experiment we have a single $v$ and a single value for $T$. This $T$ is not the same as the continuous variable $t$ that represents the passage of time.

In this case, there is no minimum of maximum, but we can still draw a graph of the function $v(T) = \frac{10 \ m}{T}$ or $T(v) = \frac{10 \ m}{v}$.

Note that this is a function that represents many different experiments, with all the possible values of $v$. In any given experiment, $v$ and $T$ are fixed (even if unknown) quantities.

Finally, if we assume that $v = 5 \ m/s$, say, then we no longer have a function and $T$ must be $2 \ s$. In this simple case, as soon as we choose either $v$ or $T$ then the function collapses to a single point on the graph. This is what happened above when you chose $v = \ 25 m\s$ as the final speed. You had then specified enough data to collapse the function relating $u$ and $T$ to a single point on the graph ($u = 75 \ m/s$ and $T = 5 \ s$). Even if you still had to calculate these numbers.

In particular, you no longer had a function $u(T)$ to do calculus on.

 

[PeroK]

But u and t are both unknown. u can be any value and t can be any value too, right?

No. Once you specify $v = 25 \ m/s$, this fixes the values of $u$ and $t$, even if you still have to calculate them.

 

[PeroK]

Hm, but how does s = ut + (1/2)at^2 take the balloon's motion into account?

Because you combine that equation for the bullet with the simultaneous equation for the motion of the balloon. If we do this formally, we have:
$$s = ut - \frac 1 2 gt^2$$and$$s_b = h_0 + v_bt$$And we set $s = s_b$ to get the equation relating the initial speed of the bullet and the time of impact:
$$uT - \frac 1 2 gT^2 = h_0 + v_bT$$This equation then contains all the information relating the initial speed of the bullet with the time of impact. Note that $h_0$ and $v_b$ are fixed quantities and not variables in this case.

Note that $u, T$ are variables here representing many different physical experiments. And we have a function relating them.

But, if we specify additionally the final speed of the bullet when it hits the balloon, then there is only one possible initial speed and one possible time. This is because if we fix $v = u -gT = 25 \ m/s$, then we have another equation relating $u$ and $T$. And the equation above reduces to:
$$u = 75 \ m/s \ \ \text{or} \ \ T = 5 \ s$$

 

[Shreya]

Essentially, the last 2 equations specify the final Velocity, thereby fixing $u$ & $T$, even though they are unknown, they refer to one specific experiment. The graph has collapsed to a single point and is no longer differentiable. The value of $u$ still has to be calculated using simultaneous equations but has already been fixed on specifying $v$. But on using $u-gT$ for $v$ I still refer to multiple experiments and multiple values of $u$ and $t$ and therefore I can use function minimization.
Please correct me if something is wrong

 

[Differentiate it]

Essentially, the last 2 equations specify the final Velocity, thereby fixing $u$ & $T$, even though they are unknown, they refer to one specific experiment. The graph has collapsed to a single point and is no longer differentiable. The value of $u$ still has to be calculated using simultaneous equations but has already been fixed on specifying $v$. But on using $u-gT$ for $v$ I still refer to multiple experiments and multiple values of $u$ and $t$ and therefore I can use function minimization.
Please correct me if something is wrong

Isn't there multiple values for u and t even though we fix v as 25?
We can still represent u as a function of t, it doesn't collapse it to a point

 

[Differentiate it]

Essentially, the last 2 equations specify the final Velocity, thereby fixing $u$ & $T$, even though they are unknown, they refer to one specific experiment. The graph has collapsed to a single point and is no longer differentiable. The value of $u$ still has to be calculated using simultaneous equations but has already been fixed on specifying $v$. But on using $u-gT$ for $v$ I still refer to multiple experiments and multiple values of $u$ and $t$ and therefore I can use function minimization.
Please correct me if something is wrong

Isn't there multiple values for u and t even though we fix v as 25?
We can still represent u as a function of t, it doesn't collapse

 

[Shreya]

Isn't there multiple values for u and t even though we fix v as 25?
We can still represent u as a function of t, it doesn't collapse it to a point

Yes, but let's not forget that the distance the balloon has to travel is fixed - and on fixing the final Velocity - the graph collapses to a single point which is (5,75)
Please correct me if I'm wrong

 

[PeroK]

Isn't there multiple values for u and t even though we fix v as 25?
We can still represent u as a function of t, it doesn't collapse

No, it's fully constrained by specifying that it impacts the balloon at $25 \ m/s$. There is only one physical solution for the initial speed ($u = 75 \ m/s$) and only one solution to the simultaneous equations.

You may be confusing the trajectory of the bullet, defined by $s = ut - \frac 1 2 gt^2$ and the relationship between the initial speed and the time at which the bullet hits the balloon: $u(T)$.

It's critical to use different symbols ($t$ and $T$) for these concepts. Using the same symbol $t$ for both only works as long as you keep the concepts separate in your mind.

 

[Differentiate it]

No, it's fully constrained by specifying that it impacts the balloon at $25 \ m/s$. There is only one physical solution for the initial speed ($u = 75 \ m/s$) and only one solution to the simultaneous equations.

You may be confusing the trajectory of the bullet, defined by $s = ut - \frac 1 2 gt^2$ and the relationship between the initial speed and the time at which the bullet hits the balloon: $u(T)$.

It's critical to use different symbols ($t$ and $T$) for these concepts. Using the same symbol $t$ for both only works as long as you keep the concepts separate in your mind.

I don't understand...
Also, does a function have no minima if there's 2(or maybe more) minimum values? Because here u is minimum, s and t are maximum. Whereas v is minimum. But when you substitute in say u -gt for v, instead of to minimum variables u and v, you have only u.
And the function u(t) has a minima
If yes, why?
It's probably wrong, but could you explain it more? I don't get how u is fixed to 75 if I set v = 25.

 

[PeroK]

I don't understand...
Also, does a function have no minima if there's 2(or maybe more) minimum values? Because here u is minimum, s and t are maximum. Whereas v is minimum. But when you substitute in say u -gt for v, instead of to minimum variables u and v, you have only u.
And the function u(t) has a minima
If yes, why?
It's probably wrong, but could you explain it more? I don't get how u is fixed to 75 if I set v = 25.

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

 

[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!