A student has plotted the graph and drawn two lines of extreme fit, one of maximum slope that still fits the points (m1) and one of minimum slope that still fits the points (m2). This is shown in the figure below.The student wants to determine the slope of m1. The two data point values which lie on m1 are of the form (x1,y1) and (x2,y2).Using the two points (0.04 m2, 0.15 N) and (0.43 m2, 1.12 N); determine the value of the gradient to two decimal places including units.
Question
A student has plotted the graph and drawn two lines of extreme fit, one of maximum slope that still fits the points (m1) and one of minimum slope that still fits the points (m2). This is shown in the figure below.The student wants to determine the slope of m1. The two data point values which lie on m1 are of the form (x1,y1) and (x2,y2).Using the two points (0.04 m2, 0.15 N) and (0.43 m2, 1.12 N); determine the value of the gradient to two decimal places including units.
Solution 1
To calculate the slope (m1) of a line, we use the formula:
m1 = (y2 - y1) / (x2 - x1)
Given the two points (0.04 m2, 0.15 N) and (0.43 m2, 1.12 N), we can substitute these values into the formula:
m1 = (1.12 N - 0.15 N) / (0.43 m2 - 0.04 m2)
m1 = 0.97 N / 0.39 m2
m1 = 2.48718 N/m2
Rounding to two decimal places, the slope m1 is 2.49 N/m2.
Solution 2
To calculate the slope of a line, we use the formula:
slope = (y2 - y1) / (x2 - x1)
Here, the two points given are (0.04 m2, 0.15 N) and (0.43 m2, 1.12 N). Let's plug these values into the formula:
slope = (1.12 N - 0.15 N) / (0.43 m2 - 0.04 m2) slope = 0.97 N / 0.39 m2
Now, we just need to calculate the division to get the slope.
slope = 0.97 N / 0.39 m2 = 2.48718 N/m2
Rounding to two decimal places, the slope of the line m1 is 2.49 N/m2.
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