1. Unit 1 Notes

2. Linearizing a Graph If a graph from a lab does not show a linear relationship, you will have to make a modification to the data to linearize the graph. Ex. t (s)Pos (m) 11 24 39 416 525

3. The graph of y vs x is an upward opening parabola, so the graph will show y vs x^2. Please note how the first column and the x-axis are labeled. t^2 (s^2)Pos(m) 11 44 99 16 25

5. Next, find the y-intercept. Start with the general form y=mx+b. In this case, y is Pos, x is t^2, and m is 1 m/s^2. Choose one point on the line and use its x- and y-values with the slope you already found. Y = mx + b Pos = m t^2 + b Using the point (25s^2, 25m), 25 = 1 *25 + b b=0

6. Now we have everything we need to finish the graph. Math Model:(the equation for your line) Pos = 1(m/s^2) t^2 Written Relationship: Position is proportional to the square of time. (found on the handout that goes on the inside of the front cover of your lab notebook) ***The written relationship does not change just because the graph has been linearized.***

7. Physical meaning of slope: The general form is: As x increases by 1 (x-unit), y increases by m (y-units). In this case, As time squared increases by 1 second squared, position increases by 1 meter.

8. Write the written relationship, physical meaning of slope, and math model on the linearized graph:

9. Meaning of Y-Intercept When x is 0, y is b. From the example, b=0. When time^2 is 0, position is 0.