Use the data and the graph to answer the following questions:
1. Write the equation of the line.
2. Give an interpretation of slope(distance, speed) and intercept(distance, speed).
3. Any insight in the predicted values.
4. Use your graph to approximate the stopping distance at a speed of 20 m/s.
I need to form a trend line; which two numbers would I use for forward and backward?
Answer (2)
With a current of 15.0 A flowing for 30 seconds, a total charge of 450 C passes through. This translates to approximately 2.81 x 10^21 electrons flowing through the device during that time. The calculation uses the relation between current, time, and charge, as well as the charge of a single electron.
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To form a trend line in a graph, you'll use the slope-intercept form of a linear equation, which is given by:
y = m x + b
where:
m is the slope of the line
b is the y-intercept
Based on the table you provided, you have data on speed (in m/s) and stopping distance (in meters). To create a trend line:
Determine the Slope (m): The slope can be found using two points from your data set. It is calculated using the formula:
m = ( x 2 − x 1 ) ( y 2 − y 1 )
For example, using the points (8, 0.61) and (14, 1.57):
m = ( 14 − 8 ) ( 1.57 − 0.61 ) = 6 0.96 = 0.16
Find the y-intercept (b): To find b , use one of the points in the equation y = m x + b . Let's use point (8, 0.61):
0.61 = 0.16 × 8 + b 0.61 = 1.28 + b b = 0.61 − 1.28 = − 0.67
Equation of the Line: Substitute the values of m and b into the equation:
y = 0.16 x − 0.67
Approximate Stopping Distance at 20 m/s: Substitute x = 20 into the equation:
y = 0.16 × 20 − 0.67 y = 3.20 − 0.67 = 2.53
So, the stopping distance at a speed of 20 m/s is approximately 2.53 meters.
Interpretation: The trend line helps us predict the stopping distance based on speed. The slope (0.16) indicates that for each additional meter per second increase in speed, the stopping distance increases by 0.16 meters. The intercept (-0.67) is the theoretical stopping distance when the speed is 0 m/s, although, in practical terms, this value may not be meaningful for real-world interpretation as stopping distance can't be negative.
