Practice Problems: Graphing Solutions
1. A car accelerates from rest along a straight line. Its velocity (v) is measured at different times (t):
t (s): 0, 2, 4, 6, 8
v (m/s): 0, 8, 16, 24, 32
Graph this data as velocity vs. time and determine the type of mathematical relationship that exists between the variables. Also, find the equation that shows the relationship.
There is a positive linear relationship between velocity and time according to this data.
An equation that would represent this data is: v = kt, where k is the slope of the best fit line.
The slope can be calculated as k = Δv/Δt
Since all the data points lie along the line of best fit, we can chose any two points to determine the slope.
k = Δv/Δt = (32 - 24)/(8 - 6) = 8/2 = 4m/s/s = 4 m/s2
The final equation is v = 4t
2. A student investigates the relationship between the pressure (P) and volume (V) of a fixed amount of gas at constant temperature:
V (Liters): 1.0, 2.0, 3.0, 4.0, 5.0
P (atmospheres): 5.0, 2.5, 1.67, 1.25, 1.0
Graph this data as pressure vs. volume and determine the type of mathematical relationship that exists between the variables. Also, find the equation that shows the relationship then use that equation to predict the pressure when the volume is 3.5 Liters.
There is an inverse relationship between pressure and volume according to this data.
An equation that would represent this data is: P = k/V, where k is constant.
Analysis of the data indicates that k = 5
The equation that fits the data is P = 5/V
When V = 3.5 L,
P = 5/3.5 = 1.4 atm
3. A student measures the period (T) of a pendulum, the time it takes for one complete swing, for different lengths (L). The data is as follows:
L (m): 0.2, 0.4, 0.6, 0.8, 1.0
T (s): 0.89, 1.26, 1.55, 1.79, 2.00
Create 2 graphs: The first is Period vs. Length and the second is Period Squared vs. Length. Identify the mathematical relationships shown in each one. What advantage does the second one have over the first? Find the equation for the line of best fit in the second graph. Use that equation to find the period when the length is 1.35 m.
The graph of Period vs. Length likely shows a quadratic relationship between the variables based on the curvature of the scatter plot. The graph of Period Squared vs. Length is linear.
The equation for the linear graph can be determine by finding the slope and the y-intercept. Chosing two points on the best fit line (as shown on the graph) allow the slope to be calculated:
slope = Δy/Δx = 3/0.5 = 6 s2/m
Extrapolation of the best fit line shows that the y-intercept is zero.
The equation of the best fit line is T2 = 6L
When L = 1.35 m
T2 = 6(1.35) = 8.1 s2
T= 2.8 s
The second graph has advantages over the first one due to being linear. It's simple to find the equation of a straight line. The second graph is a "linearization" of the first one. It's a common technique used to simplify analysis.
4. Use the graphing simulation below to find the best fit linear equation for the following unitless data:
x-data: 0, 0.8, 2.1, 3.0, 4.0
y-data: 6.8. 5.0, 3.3, 2.7, 0
Use the equation to find the y-value when the x-value is 2.6
Click both the "curve" and "value" boxes to get the best fit line and to place the data point (the orange dots) at the correct positions.
The equation determined by the simulation is y = -1.57 + 6.7
The is a negative linear relationship such that when x increases y decreases in proportion.
When x = 2.6
y = -1.57(2.6) + 6.7
y = 10.8
5.Use the graphing simulation below to find the best fit quadratic equation for the following unitless data:
x-data: 0, 2.4, 4.8, 6.8, 8.7
y-data: -0.1. 0.5, 1.2, 1.8, 3.7
Use the equation to find the y-value when the x-value is 5.2
Click both the "curve" and "value" boxes to get the best fit line and to place the data point (the orange dots) at the correct positions.
The equation determine by the simulation is y = 0.046x2
When x = 5.2 y = 0.046(5.2)2 = 1.2