# A new take on a relative velocity problem

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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##.

• Shreya
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.

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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.

• Shreya
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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##!

• Shreya
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 🙏

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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.

• Shreya and Differentiate it
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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.

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

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@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 🙏

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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.

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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##.

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• Shreya
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.

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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?

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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

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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".

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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

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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

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• Shreya