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

[Astro-Eddie]

TL;DR

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...