APPHYSICS

Part 3: Linearization How do we create a mathematical model for a line that isn’t straight? For example, a student wanted to experimentally determine the relationship between a circle’s area and its radius. Here is the data that student collected. radius (cm) area (cm 2 ) 5.0 78.5 10.0 314.2 15.0 706.9 20.0 1256.6 25.0 1963.5 5. Graph the data on the graph below and draw a line of best fit. Make sure to plot diameter on the x-axis and circumference on the y-axis. Don’t forget to label your axes. 6. What shape is the line? Hopefully you noticed that the line wasn’t straight. It is difficult to get the equation for a line that isn’t straight unless you have a computer. Or unless you know how to use linearization, which is the process of transforming a curvy line into a straight line so that you can analyze it. How does that work?

Part 4: Interpreting Graphs: In physics, the area under a graph or the slope of a graph can hold meaningful information. In this section, we will focus on how to find the area under the graph and identify units for the area under the graph. Additionally, we will review how to find slope and determine units for a given slope. Example 1: First, let’s describe and determine the slope  Description of the slope:  The slope is CONSTANT and POSITIVE.  The Y-INTERCEPT is 5m/s, meaning the starting speed is 5m/s  Using the slope equation of: m = y 2 − y 1 x 2 − x 1 we can determine the slope of the line to be m = 8 m − 6 m 3 s − 1 s = 1. Since the units on the bottom of the equation are s, seconds and the units on the top of the equation are m/s, meters per second the unit of the slope would be 1 m s s ∨ 1 m s 2 Now, let’s look at the area under the graph.  The area MUST first be broken down into geometry that we can solve for! The area under the line can be broken down into a triangle and a rectangle as indicated by the shapes outlined on the graph. Velocity (m/s) Time (s) Velocity (m/s) = 5m/s Time (s) = 5s Time (s) = 5s Change in Velocity (m/s) = FROM 5m/s to 10m/s or a height of 5m/s To determine the Area of a Triangle = 1 2 b ∗ h  H - Change in Velocity =10m/s - 5m/s = 5m/s  B – Time = 5s o 1 2 b ∗ h = 1 2 ( 5 s ) x ( 5 m s ) ¿ 12.5 m ∗ s→ 12.5 m To determine the Area of a Rectangle = Lx W For our example, instead of L x W we will use H x W  H - Velocity = 5m/s  W – Time = 5s o H x W = 5 m s x 5 s = 25 m s ∗ s→ 25 m o AREA = DISTANCE

5) mgh = 1 2 k x 2 (solve for k) 6) v x 2 = v xo 2 + 2 a x ( x − x o ) (solve for x o )