How to predict the theoretical slope, [M], and y-intercept, [B] on your graphs?

Example 1: Graph “v versus t” in Experiment 2, PHYS2LA, or PHYS40A

Usually, Graph “Y versus X”

So, we treat v as Y and t as X.

We also have v - gt = 0. These equations are usually given or you should have known them from PHYS2 or PHYS40 series.

We may re-write it as Y - [M]X = [B].

Then, Y = [M]X + [B].

Slope: [M] = (D Y/D X) = (dY/dX).

In this case, [M] = g.

It is so-called the first-order differential of Y with respect to X. You can find it in every Calculus textbook.

Y-intercept: [B] = Y value @ X = 0.

In this case, [B] = 0.

It is from Algebra. If you do not understand it, review your Algebra textbook.

Example 2: Graph "m versus T²" in Experiment 1, PHYS2LC, or Experiment 2, PHYS40B

Usually, Graph “Y versus X”

So, we treat m as Y and T² as X.

We also have  m = (k/4)as Y and T² + M/3 as X- gt = 0. These equations are usually given or you should have known them from PHYS2 or PHYS40 series.

We may re-write it as Y - [M]X = [B].

Then, Y = [M]X + [B].

Slope: [M] = (D Y/D X) = (dY/dX).

In this case, [M] = g.

It is so-called the first-order differential of Y with respect to X. You can find it in every Calculus textbook.

Y-intercept: [B] = Y value @ X = 0.

In this case, [B] = 0.

It is from Algebra. If you do not understand it, review your Algebra textbook.
 
 

Example 3: Graph "1/s’ versus 1/s" in Experiment 6, PHYS2LC

Usually, Graph Y versus X

So, we treat 1/s’ as Y and 1/s as X.

We also know that (1/s’) + (1/s) = (1/f).

We may re-write it as Y + X = (1/f).

Then, Y = -X + (1/f).

Slope: [M] = (D Y/D X) = (dY/dX).

Here, [M] = -1.

Y-intercept: [B] = Y value @ X = 0

Here, [B] = (1/f).

Make sure you understand this skill to make a theoretical prediction because that definitely you will encounter it more than once during PHYS2L or PHYS40 series.

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