Question 1

Mathematical Physics

Question 1 – Where do these equations come from?

The first approach I am going to make is from an underlying assumption.

Look around you or read a physics book lying beside you – gravity is here.

It is the acceleration that pulls you to the ground every day. Thus we can

begin by assuming acceleration equals gravity when an object is in the air.

and remember

Therefore and it follows that

Now we have (where c is equal to the initial velocity or )

Since then it follows that

And after pushing it through the integration machine we now have

As for “c” in this second case when time equals zero “x” will equal “c” and thus “c” is the initial position or .

So finally by our first method of attack we have .

Since any variable behaves the same as x we have a second equation .

A second approach to prove the same equations with the same beginning and end takes down a road less traveled – the geometric proof.

Here we have a representation of what the acceleration (gravity) vs. time graph looks like

The units on the y-axis are those of acceleration (m/s/s) and those of the x-axis are time (s). Notice if we multiply the units together

we get velocity (m/s) which is also the area under the curve. Thus we take the equation of y (g) and multiply it times the general time (t)

so that we can define the velocity at any given time. Thus we have v = gt.

Here we have an interesting predicament.

This is a graph of the velocity vs. time if the initial velocity is zero (v = gt).

To find distance from this graph we will multiply the units on the y-axis (velocity – m/s) and the x-axis (s).

By using the triangle area formula to get distance (m) in terms of time (s). Since the y axis is height and the x axis is the

Base then . In this particular case and so therefore

But if the initial isn’t zero but is a constant (we will use 3 m/s arbitrarily) then we have different graph.

The process continues from the previous example but we have to account for the area of the rectangle produced by the initial velocity.

The area of this rectangle is simply the initial velocity times time. . Now if we add the two together we get

Here is a graph of the last equation -

There is a difference between this graph and the next…

It’s the y-intercept. Since units for the y-axis are meters then this y-intercept is the initial location of the object in other words the equation must incorporate

an initial position ().

Hereby we have the final equation

A sub question to address in this section is, “What will distinguish these two equations?”

This is where our proportion givers come in – the famous sine and cosine!

Thus far we have assumed gravity as the initial acceleration for both equations and for the initial velocity in both situations. But there is a difference. Since velocity is a vector in two-dimensional space it will have two components – one for the x direction and one for the y direction.