B Where does the 1/2 in 1/2 at^2 come from?

[Astro-Eddie]

TL;DR Summary

Why is distance=1/2 at^2? Can you explain without using calculus? (I'm not in B. I'm lower)

I am currently studying Newton's laws and mechanics. I have this question: Why is distance=half a*t^2? Where did the 1/2 come from? Can someone explain this without using calculus?

 

[berkeman]

TL;DR Summary: Why is distance=1/2 at^2? Can you explain without using calculus? (I'm not in B. I'm lower)

I am currently studying Newton's laws and mechanics. I have this question: Why is distance=half a*t^2? Where did the 1/2 come from? Can someone explain this without using calculus?

No, sorry. The equation comes directly from the calculus involving integration and differentiation. How long until you take your first calculus class?

https://en.wikipedia.org/wiki/Equations_of_motion

 

[gneill]

A very simple explanation would be if you start from velocity = 0 and apply a constant acceleration, then if you plot velocity versus time it would form a triangular shape between the plot and the time axis.

The area under the curve is equal to the distance covered. What's the area of that triangle?

 

[Baluncore]

Consider a constant acceleration from zero.
The area of the acceleration graph with time is velocity; v = a * t ;
The velocity graph is therefore a triangle against time.
The area of a triangle is = ½⋅height⋅base
The area under the velocity graph against time is displacement; s.
s = ½⋅v⋅t = ½⋅a⋅t² .

 

[berkeman]

(must.resist.urge.to.remind.that.areas.are.coming.from.integrations...)

 

[topsquark]

TL;DR Summary: Why is distance=1/2 at^2? Can you explain without using calculus? (I'm not in B. I'm lower)

I am currently studying Newton's laws and mechanics. I have this question: Why is distance=half a*t^2? Where did the 1/2 come from? Can someone explain this without using calculus?

Say that we have an object that starts at time $t_0 = 0$ s, at a point $x_0$, with an initial velocity $v_0$ and moves with a constant acceleration $a$, and travels for a time $t$. (All in 1D.)

We know that
$v = v_0 + at$

and
$\overline{v} = \dfrac{x - x_0}{t}$ $\leftarrow$ average velocity

Now, the average velocity can also be found as
$\overline{v} = \dfrac{v_0 + v}{2}$
(for motion with constant acceleration.)

So let's put some of this together:
$\dfrac{x - x_0}{t} = \dfrac{v_0 + v}{2}$

or
$x = x_0 + \dfrac{v_0 t}{2} + \dfrac{v t}{2}$

and using $v = v_0 + at$

$x = x_0 + \dfrac{v_0 t}{2} + \dfrac{(v_0 + at) t}{2}$

$x = x_0 + \dfrac{v_0 t}{2} + \dfrac{v_0 t}{2} + \dfrac{1}{2} a t^2$

$x = x_0 + v_0 t + \dfrac{1}{2} a t^2$

-Dan

 

[Nugatory]

Easiest explanation:

We start at speed zero, we accelerate for time $t$ so our speed is $at$.
Our average speed therefore is the average of $0$ and $at$, which is $\frac{1}{2}at$.
(This only works for constant acceleration, but that's we're doing here)

Distance traveled is speed times time, so If we move at an average speed of $\frac{1}{2}at$ for time $t$ the distance is $\frac{1}{2}at^2$.

Take a moment to draw a rectangle of height $\frac{1}{2}at$ and width $t$, compare with the triangle that @gneill drew in post #3 above.

 

[Astro-Eddie]

Say that we have an object that starts at time $t_0 = 0$ s, at a point $x_0$, with an initial velocity $v_0$ and moves with a constant acceleration $a$, and travels for a time $t$. (All in 1D.)

We know that
$v = v_0 + at$

and
$\overline{v} = \dfrac{x - x_0}{t}$ $\leftarrow$ average velocity

Now, the average velocity can also be found as
$\overline{v} = \dfrac{v_0 + v}{2}$
(for motion with constant acceleration.)

So let's put some of this together:
$\dfrac{x - x_0}{t} = \dfrac{v_0 + v}{2}$

or
$x = x_0 + \dfrac{v_0 t}{2} + \dfrac{v t}{2}$

and using $v = v_0 + at$

$x = x_0 + \dfrac{v_0 t}{2} + \dfrac{(v_0 + at) t}{2}$

$x = x_0 + \dfrac{v_0 t}{2} + \dfrac{v_0 t}{2} + \dfrac{1}{2} a t^2$

$x = x_0 + v_0 t + \dfrac{1}{2} a t^2$

-Dan

is the v in the very first equation the final velocity? Also, why d we divide Vo+V by 2 for the average velocity?

 

[topsquark]

is the v in the very first equation the final velocity? Also, why d we divide Vo+V by 2 for the average velocity?

Yes, v is the final velocity.

As to the average, for example, how do you find the average of the set
(v,t) = {(2.0,0), (2.5,1), (3.0,2), (3.5,3), (4.0,4)}?

That would be the middle point of the set, so
$\left ( \dfrac{2.0 + 4.0}{2}, \dfrac{0 + 4}{2} \right ) = (3.0, 2)$ (in whatever units.)

But this means that the average v is just the sum of the two endpoints divided by 2.

-Dan

 

[nasu]

The fact that you avoid the word "calculus" does not mean that you are not using it. You are using results from calculus when you say that the average velocity for constant acceleration is the average of the velocities at the end of the interval. It is not a self evident truth. The same for using the geometrical method. Arhimedes used it and even though he did not call it calculus, it was.
So it is true what @berkeman said, you are still using calculus to justify the factor of 1/2.

I am not saying that it is not useful to show some sort of justification for students in non-calculus physics classes. You can easily justify the fact that the distance is not $at^2$ but a smaller fraction of this, due to the fact that the velocity was smaller than the final velocity during the motion. But the fact that the fraction is 1/2 and not 1/3 or any other requires the methods that we call "calculus". Whatever method you use to "justify" it, I think it is fair to tell the students that the result can be proven by mathematical methods they will learn later in calculus.

 

[topsquark]

The fact that you avoid the word "calculus" does not mean that you are not using it. You are using results from calculus when you say that the average velocity for constant acceleration is the average of the velocities at the end of the interval. It is not a self evident truth. The same for using the geometrical method. Arhimedes used it and even though he did not call it calculus, it was.
So it is true what @berkeman said, you are still using calculus to justify the factor of 1/2.

I am not saying that it is not useful to show some sort of justification for students in non-calculus physics classes. You can easily justify the fact that the distance is not $at^2$ but a smaller fraction of this, due to the fact that the velocity was smaller than the final velocity during the motion. But the fact that the fraction is 1/2 and not 1/3 or any other requires the methods that we call "calculus". Whatever method you use to "justify" it, I think it is fair to tell the students that the result can be proven by mathematical methods they will learn later in calculus.

Whereas it is true that you can easily prove the average velocity formula using mean value integration formula, that does not mean you can't do it without Calculus. The simple example I used above proves it without invoking any Calculus.

I don't mean to say that using Calculus for Introductory Physics isn't easier, nor am I saying that none of Introductory Physics relies on it. Far from it. But projectile motion under constant acceleration, in particular, does not require Calculus in order to derive most of its results. The position formula is one such that does not require Calculus. To say that you can't understand the formulas without Calculus simply isn't true and IMHO just makes the field needlessly obscure to a new learner who does not yet know it.

Time enough to learn these concepts when they become necessary, and you really do not need Calculus for Physics I unless you are going to continue onto Physics II (where I agree that Calculus is absolutely necessary.)

-Dan

 

[nasu]

You did not prove that the average velocity (defined as total displacement over total time) is equal to the arithmetic mean of the velocities at the ends of the interval. Why should be the value in the middle of the set be equal to the average velocity defined in the usual way for motion? What is the distance travelled associated with this set of (v,t) values?

 

[Meir Achuz]

Learn calculus before going any further. The long derivation without calculus shows you why

 

[PeroK]

The fact that you avoid the word "calculus" does not mean that you are not using it.

We don't need calculus to calculate the area of a rectangle or triangle. The fact that it can also be done using calculus, doesn't mean calculus is necessary.

What is the distance travelled associated with this set of (v,t) values?

The displacement is the area under a velocity time graph. Which, for constant acceleration, is a combination of rectangles and triangles. No calculus is needed to calculate the average velocity in this case.

 

[haushofer]

For constant acceleration, the average velocity is midway between the starting and ending velocity. The distance traveled is the average velocity times elapsed time.

 

[PeroK]

We don't need calculus to calculate the area of a rectangle or triangle.

The irony is that to develop the integral calculus the area of rectangles and triangles is normally used. So, an appeal to calculus to prove that the area of a triangle is 1/2 x base x height is a circular argument.

 

[nasu]

We don't need calculus to calculate the area of a rectangle or triangle. The displacement is the area under a velocity time graph.

It's not calculating the area but the fact that the distance travelled is equal to the area. I consider this as part of calculus. The area under the curve of the graph of a function is the integral of the function. Does not matter how you calculate that integral, by measuring the area or by analytics methods.

 

[Baluncore]

Does not matter how you calculate that integral, by measuring the area or by analytics methods.

The analytics method is called calculus. Geometry predates calculus by a couple of millennia. You do not need calculus to do geometry, or work out the area of a geometric figure.
Prior to calculus, you could have solved the problem with geometry.

 

[topsquark]

You did not prove that the average velocity (defined as total displacement over total time) is equal to the arithmetic mean of the velocities at the ends of the interval. Why should be the value in the middle of the set be equal to the average velocity defined in the usual way for motion? What is the distance travelled associated with this set of (v,t) values?

What more proof do you need? Let's take this down a step. Say we're talking about motion with constant velocity. What is the average displacement for a data set for motion with a constant velocity? Is this not
$\overline{x} = \dfrac{x_0 + x}{2}$

It's simply finding the "center" of a geometric construction. Yes, if acceleration weren't constant we couldn't do this and the non-Calculus derivation of
$x = x_0 + v_0 t + \dfrac{1}{2} a_0 t^2 + \dfrac{1}{6} j t^3$

(motion with a constant jerk) would be... annoying. And, certainly, if nothing about the motion was constant you would definitely need to use Calculus. All I'm saying is that, for the specific case of constant acceleration, we don't need Calculus.

-Dan

 

[vanhees71]

The very fact that the velocity is given by the area under the curve $(t,a(t))$ is an application of the fundamental theorem of calculus. You may argue without explicitly writing integrals, but you'll have to argue with arguments leading to the definition of the integral (most "naturally" using the Riemann integral in this case). I never understood, what's the merit of this "calculus-free approach" to physics. You don't need the full formalism of 19th-century analysis to get pretty far in physics, but introducing the notion of derivatives and integrals at least on this heuristic level, makes things way more clear than avoiding them.

 

[PeroK]

The very fact that the velocity is given by the area under the curve $(t,a(t))$ is an application of the fundamental theorem of calculus.

The proof of which rests on the known area of rectangular and triangular shapes. These simple geometric calculations predate calculus, as mentioned above.

 

[vanhees71]

The fact to derive that it IS this area is already calculus.

I'd start with the definition of "momentary velocity" in the usual way to calculate finite differences $\Delta \vec{x}/\Delta t$ and take the limit $\Delta t \rightarrow 0$. That's of course the derivative, and one should be allowed to call it such and also to derive the basic rules how to calculate with it (taking derivatives of the elementary functions, product/quotient rule etc.).

Analogously you get the acceleration as the time derivative of velocity.

Then that the opposite operation is integration and leads to the "curve under" the velocity-component-time diagram to get the corresponding position-vector component can also be motivated in a pretty intuitive geometric way (analogously the velocity from a given acceleration $a(t)$).

All that's needed from "elementary measure theory" is indeed that the area of a rectangle is length times width.

 

[nasu]

The analytics method is called calculus. Geometry predates calculus by a couple of millennia. You do not need calculus to do geometry, or work out the area of a geometric figure.
Prior to calculus, you could have solved the problem with geometry.

Which problem would you have solved with geometry? How do you know what area to calculate to get the distance travelled in a non-uniform motion? What geometry theorem tells you this?

 

[nasu]

The fact to derive that it IS this area is already calculus.

Yes, this is what I mean. The problem is not not how to calculate the area.

 

[hutchphd]

TL;DR Summary: Why is distance=1/2 at^2? Can you explain without using calculus? (I'm not in B. I'm lower)

Can someone explain this without using calculus?

No, sorry. The equation comes directly from the calculus involving integration and differentiation. How long until you take your first calculus class?

You did not prove that the average velocity (defined as total displacement over total time) is equal to the arithmetic mean of the velocities at the ends of the interval.

The irony is that to develop the integral calculus the area of rectangles and triangles is normally used.

The very notion of an intantaneous velocity and acceleration is at the heart of differential calculus ) so the OP is self-referential, leading us directly to wallow in the semantic swamp. Surf's up...

 

[haushofer]

The proof of which rests on the known area of rectangular and triangular shapes. These simple geometric calculations predate calculus, as mentioned above.

I'm building a working office in my house right now. I had to determine the area of a rectangle to buy the floor. My construction worker told me to apply calculus.

 

[haushofer]

My mother needed tiles for her garden. She wants a nice triangle and asked me how many tiles she should buy. I told her to use calculus.

 

[haushofer]

What's the area of a football field you ask? Worry not. Use calculus.

 

[haushofer]

What I try to say is you shouldn't add apples and oranges. Even though that's calculus.

 

[berkeman]

My mother needed tiles for her garden. She wants a nice triangle and asked me how many tiles she should buy. I told her to use calculus.

Does she still love you?

Good thing she didn't ask you to optimize something...

 

[topsquark]

Folks, I'll start this by saying I'm on a new medication and that I'm a little b!tchy. But please consider the following:
1. The OP requested a non-Calculus proof for the formula. This implies that they either have not taken Calculus or that they are in a non-Calculus course.

2. Mentioning that Calculus makes Physics easier is fine. Saying that it is a requirement in order to solve this problem is simply not true.

3. Telling the OP that you need Calculus to understand where the equation comes from is wrong, irrelevant to the OPs needs, and somewhat elitist. I totally agree that Calculus is an essential tool for understanding Physics. But it is not what everyone needs.

Please consider that this whole conversation about whether Calculus is necessary to understand Physics, or that Calculus makes it easier, is completely outside of the question posed by the OP. There is already another thread in progress about "how can you teach Physics without Calculus." I recommend that you put your arguments there, where it is more to the point, rather than here, where it is simply not relevant.

-Dan

 

[nasu]

If the OP needs an explanation without using some tool, the need is supposed to make it happen even if there isn't one?

He can get a justification, an intuition for why may be so, for why is reasonable to accept it for now as it is. Is done all the time in high school physics and algebra based university courses. What is the harm in telling him that this is not a proof without calculus, just a proof that uses an easier to understand method or result of calculus.

Like, area under the graph of force versus distance is equal to the work of the force or area under the graph of velocity versus time is the distance travelled. By using this he can see why for uniform acceleration the average velocity is equal to the arithmetic mean of velocity and why isn't in general so. Or why the work of the elastic force is the way it is.

But this not "without calculus", just without the word "calculus". Kind of politically correct physics in algebra based courses where the students get offended if they hear "calculus".

 

[topsquark]

If the OP needs an explanation without using some tool, the need is supposed to make it happen even if there isn't one?
He can get a justification, an intuition for why may be so, for why is reasonable to accept it for now as it is. Is done all the time in high school physics and algebra based university courses. What is the harm in telling him that this is not a proof without calculus, just a proof that uses an easier to understand method or result of calculus.
Like, area under the graph of force versus distance is equal to the work of the force or area under the graph of velocity versus time is the distance travelled. By using this he can see why for uniform acceleration the average velocity is equal to the arithmetic mean of velocity and why isn't in general so. Or why the work of the elastic force is the way it is.
But this not "without calculus", just without the word "calculus". Kind of politically correct physics in algebra based courses where the students get offended if they hear "calculus".

But the derivation does not require Calculus! The OP didn't want a Calculus based explanation, and it did not need one. My reply essentially comes from my High School Physics class, which was not an AP course (my High School didn't offer AP.) It is correct in all the details, and doesn't require any Calculus to understand it.

Yes, the proof is 4 lines shorter if you use Calculus. So what? The question is not about teaching the OP about areas under the curve nor is it about nor why the arithmetic mean of the velocity is not the same as the average velocity in the general case. When I presented my proof I did mention that the arithmetic mean equation only worked for constant acceleration. But by insisting on mentioning Calculus you are moving directly away from what the OP was asking: How to show where the 1/2 comes from without using Calculus. You aren't answering the question: you are expanding on it. That is simply not necessary in this case and it can only serve to put off the OP.

Yes, some people are threatened by Calculus. I don't really know why. Personally, I found my Calculus classes to be relatively easy. (So easy that I effectively taught myself most of Calculus I - III while I was in my Senior year of High School, mistakenly thinking that the whole text was just Calculus I.) But the fact remains that some people are. So why pressure the point? If the student is not bound to be a Physics, or related, major why worry about if they are not told a Calculus based proof when a non-Calculus based proof was readily available?

Just to be clear: whereas I am saying that it is possible to show where the 1/2 comes from without Calculus, I am generally not against your comments as to how Calculus is needed in learning Physics properly. It's just that it doesn't sound like the OP needs to learn those lessons, thus I do not believe that they belong in this thread: the Calculus comments are a tangent and do nothing to help the OP.

-Dan

 

[hutchphd]

the Calculus comments are a tangent and do nothing to help the OP.

If the OP thinks that his statement of the problem does not already use calculus he needs to be educated. I presume this need is why he has self-identified as a student. The very notion of an instantaneous velocity is the crux: an educator needs to be certain that knowledge is part of his Physics education despite the demand of "no calculus". The only question is how advanced.
I wish to learn English Literature but no Shakespeare.......
American History but no slavery please. I am weary.

 

[Vanadium 50]

I think we're getting tangled up in knots here.

Let's review - the OP asked where the 1/2 came from, and was told that the distance traveled is the average speed times the time. The average is half the sum of the initial and final velocities.

This is a fine, non-calculus based answer. Unfortunately, the OP rejected it.

Now we're trying to defend it by coming up with some sort of hand-wavy crypto-calculus. Given that we will never get to the bottom of the "but why?" ladder, nor can we guess what answer the OP finds acceptable, maybe we should have stopped there.

 

[Astro-Eddie]

Yes, v is the final velocity.

As to the average, for example, how do you find the average of the set
(v,t) = {(2.0,0), (2.5,1), (3.0,2), (3.5,3), (4.0,4)}?

That would be the middle point of the set, so
$\left ( \dfrac{2.0 + 4.0}{2}, \dfrac{0 + 4}{2} \right ) = (3.0, 2)$ (in whatever units.)

But this means that the average v is just the sum of the two endpoints divided by 2.

-Dan

Thank you so much!

 

[topsquark]

If the OP thinks that his statement of the problem does not already use calculus he needs to be educated. I presume this need is why he has self-identified as a student. The very notion of an instantaneous velocity is the crux: an educator needs to be certain that knowledge is part of his Physics education despite the demand of "no calculus". The only question is how advanced.
I wish to learn English Literature but no Shakespeare.......
American History but no slavery please. I am weary.

Where, exactly, did the OP ask about instantaneous velocity? And at what point are we designated as the educators in his/her course that we should teach Calculus in what is obviously a non-Calculus Physics course?

The OP asked a specific question. That question was answered in the manner that the OP asked for. Why are we supposed to add "...but the OP needs to know that Calculus is necessary to learn Physics" when the OP has clearly asked for a non-Calculus explanation, and say (incorrectly) that Calculus is necessary to understand where the 1/2 comes from?

To continue your example, if someone is asking a question about "The Metamorphosis" do we really need to say that Caliban wanted to rape Miranda so that he could create a race of Calibans on his island? Or that Prospero stopped him? Or, if someone is asking a question about the Maginot line, do we need to say that the Emancipation Proclamation was not what ended slavery in the US and that it was actually the 13th Amendment that legally made slavery a crime?

If someone asked a question about the force exerted by one charged particle on another, do we need to tell them that the Coulomb force is conservative and that they need to know how to prove that? Why not just answer the question?

Honestly, I'm not trying to tell anyone what they can/should say in a thread: I'm not Staff and I don't set the site policies. But the whole conversation about the need for Calculus in order to fully understand Physics is completely extraneous to the question that was asked. Are we here to answer questions that were posed or, seriously, are we here to tell the members that they don't know enough to answer questions that they didn't ask about? Sometimes, I agree, that comment does need to be made, but it did not need to be made here.

-Dan

 

[vanhees71]

I think the teaching of physics should also provide an idea, where the formulae come from you use to solve problems. In fact it's the real subject of physics. It's not physics to put numbers in a given formula, which you don't understand.

The very notion of the kinematical quantities needs calculus, whether you are allowed to call it calculus or not. The "calculus-free approach" is making things much less transparent, and I've no clue where the idea comes from that one should avoid it at all costs. As stressed several times, the notion that position-vector components can be calculated as the area under the space-time diagram graph (up to an additive constant, which is given by the initial conditions) already needs a heuristic idea of calculus. Then of course for $v_j=a_j t$ with $a_j=\text{const}$ this area of a triangle is given by the elementary geometric definition. The point is to understand why that's the case, and for this you need a heuristic understanding of calculus/integration.

 

[PeroK]

I think the teaching of physics should also provide an idea, where the formulae come from you use to solve problems. In fact it's the real subject of physics. It's not physics to put numbers in a given formula, which you don't understand.

The very notion of the kinematical quantities needs calculus, whether you are allowed to call it calculus or not. The "calculus-free approach" is making things much less transparent, and I've no clue where the idea comes from that one should avoid it at all costs. As stressed several times, the notion that position-vector components can be calculated as the area under the space-time diagram graph (up to an additive constant, which is given by the initial conditions) already needs a heuristic idea of calculus. Then of course for $v_j=a_j t$ with $a_j=\text{const}$ this area of a triangle is given by the elementary geometric definition. The point is to understand why that's the case, and for this you need a heuristic understanding of calculus/integration.

Where does calculus come from? If you insist on teaching calculus before you can calculate the area of a triangle or study conic sections - we are dealing with a parabola here - then someone else might insist on a course in real analysis before you can study calculus. What's the point of using integration if you can't prove where the formulas came from?

 

[vanhees71]

I do NOT insist of teaching calculus first. I insist on giving a heuristic explanation! Of course you start with the definition of the area of rectangles as (width times hight) of plane geometry. Then you need the idea of calculus to calculate the area under the graph of curve by first defining it approximately by the sum of areas of rectangles, which you make ever smaller to get a more and more accurate value for this area.

To get from $v_j$ to $x_j$ (up to an additive constant) again you need that $v_j=\dot{x}_j$, defined by the limit of finite differences $\Delta x_j/\Delta t$ for $\Delta t \rightarrow 0$. Then you can get the other way by approximating $\Delta x_j$ by $v_j \Delta t_j$ and sum over these rectangles to see that $x_j$ is (up to an additive constant) given by the area under the velocity-time diagram. Then, of course, if you know this area from elementary geometric considerations like the triangle for uniform acceleration, then you can write down this area without using formal integrals.

This kind of heuristic arguments are even a better approach to the application of calculus in physics than to just teach it as analysis a la Bourbaki. Of course, the serious theoretical physicist also needs this formal approach if it comes to the finer details, but at the level of high-school physics that would of course be overdoing it. Even in the math classes you get only to the most simple beginnings of the rigorous approach, as using the "$\epsilon$-$\delta$ definition" of limits, continuity, differentiability, and all that.

 

[haushofer]

Does she still love you?

Good thing she didn't ask you to optimize something...

Yes. My mother doesn't differentiate between me and my brothers.

 

[topsquark]

If the student was in a Calculus based class, then I would agree that teaching limits, area under the curve, etc. makes sense. The OP clearly is not in such a class. (I mean, not even the Calculus proponents felt the need to mention that $x = x_0 + v_0 t + (1/2) a_0 t^2 + \dots$ is the first few terms of a Taylor (technically Maclaurin) expansion, which is a more direct approach to the 1/2 coefficient than talking about areas under the curve.) Would you insist on speaking to a resident of Venice in Latin, as opposed to his language of Italian? All I am saying is that we need to talk to the student in the language they understand. We are not here to teach the OP their whole class, just to answer the question that was asked. Some extra information is fine, but insisting that the question cannot be answered without a knowledge of Calculus is not only a bit elitist, but actually wrong. (Those that disagree should probably go back and take another look at Pythagoras, the man who invented triangles. )

Please let me just say one last time that if someone is asking a question on the site, please focus on the following when answering:
1. If possible, speak the language the OP will understand.

2. If possible, answer the question directly. If it does require a bigger outlook than the student has presented then, yes, go ahead and say that. But if it does not, consider that the bigger outlook might do nothing but confuse the student.

3. If there is a simple way to answer the question, that's likely to be the better way of doing it, unless it's clear that the student will benefit from the more general approach.

Most that come here are likely to be Physics (or related) majors and will want a more general treatment. On the other hand, not everyone that does come here is a Physics major or even wants to be one.

Okay, I'm going to leave the conversation because some of you simply will not listen to what I think is reason. I'm sure that those same some of you feel the same way about me. In any event, however you look at it, a protracted argument on this does not serve anyone. The OP seems to be satisfied by the solution presented and I don't see that anything useful will happen from here on, so I'm done.

-Dan

 

[hutchphd]

you simply will not listen to what I think is reason

This is why we should listen to each other's ideas. It forms the fundamental purpose for an honest discussion.
I do find your accusations reflexive (as you infer before the fact). But making up one's mind before the discussion starts will likely shorten it.

 

[vanhees71]

If the student was in a Calculus based class, then I would agree that teaching limits, area under the curve, etc. makes sense.

You can't teach mechanics without these ideas. To avoid to call this calculus and the adequate mathematical language for formulating physics, is nonsense, because it makes the subject more difficult rather than simpler.

The OP clearly is not in such a class. (I mean, not even the Calculus proponents felt the need to mention that $x = x_0 + v_0 t + (1/2) a_0 t^2 + \dots$ is the first few terms of a Taylor (technically Maclaurin) expansion, which is a more direct approach to the 1/2 coefficient than talking about areas under the curve.) Would you insist on speaking to a resident of Venice in Latin, as opposed to his language of Italian? All I am saying is that we need to talk to the student in the language they understand. We are not here to teach the OP their whole class, just to answer the question that was asked. Some extra information is fine, but insisting that the question cannot be answered without a knowledge of Calculus is not only a bit elitist, but actually wrong. (Those that disagree should probably go back and take another look at Pythagoras, the man who invented triangles. )

Please let me just say one last time that if someone is asking a question on the site, please focus on the following when answering:
1. If possible, speak the language the OP will understand.

If you want to learn physics you have to learn the language that is used to talk about it. If you don't want to learn, we can't help.

2. If possible, answer the question directly. If it does require a bigger outlook than the student has presented then, yes, go ahead and say that. But if it does not, consider that the bigger outlook might do nothing but confuse the student.

3. If there is a simple way to answer the question, that's likely to be the better way of doing it, unless it's clear that the student will benefit from the more general approach.

You should make things as simple as possible but not simpler. One should be aware that a topic is remembered best in the way it is taught first, and it is difficult to unlearn wrong explanations. Fortunately also didactics makes some progress with the years. The invention of "calculus-free physics" by didacticians is a misguide!

 

[weirdoguy]

To avoid to call this calculus and the adequate mathematical language for formulating physics, is nonsense,

Stop calling what I, and others, do for a living a nonsense. I've also been thought that way in high-school, and there has been hundreds of exercises that were very difficult, and really using calculus wouldn't help. See for example physics olimpiads. You don't teach in high-school, so you simply do not know what you are talking about. You don't know what struggles students have, and 99% of them would not be resolved by using calculus.
Even I still sometimes have truggles with some of the problems, and I know calculus perfectly (I teach it also).I have an exercise for you: we have a rabbit that sits in the distance $d$ frome the road. There is a car on the road moving with the speed $u$. Rabbit wants to cross the road and the car is in the distance $l$ from the rabbit when rabbit starts to move. What is the minimal speed of the rabbit so that it won't be hit by a car? At what angle should it move?

Solve it without calculus. I would say that solving it this way requires a "little bit" more of understanding physics than using calculus.

 

[vanhees71]

I do not say that you cannot solve some kinds of problems without calculus. I say that physics gets unnecessarily complicated by avoiding the adequate language it is formulated in, which is calculus.

The problem is very vaguely formulated. Should the rabbit move in with constant velocity along a straight line? Then it's solving a system of linear equations with some parameters, which for given angle and speed of the rabbit admittedly is not calculus.

I don't say that there aren't very challenging problems which do not need calculus for their solution. I also know that teachers are obliged to follow study plans that are not optimal. In Germany we also have all kinds of "experiments" in other subjects and in elementary school. The result is that 1/5-1/4 of the fourth-graders (end of elementary school in Germany) are not able to read and comprehend simple texts...

 

[hutchphd]

Solve it without calculus. I would say that solving it this way requires a "little bit" more of understanding physics than using calculus.

With respect, if you are attempting to make a point of logic here it is lost on me. Perhaps I need some help.
The point of my objection to the OP is not doctrinal. I have taught freshmen both calculus-based and not calculus-based courses. To not be allowed to utter the word "calculus" is silly. Much of the subject, regardless of semantics, deals with instantaneous rates of change. The tapdance required not to call it calculus is exhausting. Avoiding the tapdance does not require introduction of partial diff. eq. into the syllabis. Also for understanding the historical import of Newton in the scientific revolution a descriptive knowledge of where calculus fits is important. Call it fluxions if you must.

 

[Tom.G]

This thread has reminded me of a circular firing squid.

 

[hutchphd]

In the limiting case only.....