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# A graphic representation of the process of calculating the slope of a regression line from the results of time trials of 50, 100, 200, and 400 m is shown in the figure. Plotting the times versus the...

A graphic representation of the process of calculating the slope of a regression line from the results
of time trials of 50, 100, 200, and 400 m is shown in the figure. Plotting the times versus the distances
for those time trials produces a slope that expresses the critical velocity as 1.511 m/sec.
The slope of that regression line defines the expected change in time for each change in distance. In
other words, it represents the average number of meters covered during each second of swimming
for distances between 50 and 400 m. The slope of that regression line is, therefore, equal to the critical
swimming velocity. That velocity can be represented as a time per 100 m simply by dividing it into 100.
The question is, how was this critical velocity of 1.511 m/sec calculated from these values?
50 m = 28.90 sec or 1.73 m/sec
100 m = 61.35 sec or 1 .63 m/sec
200 m = XXXXXXXXXXsec or 1.50 m/sec
400 m = XXXXXXXXXXsec or 1.42 m/sec
Answered Same Day Jun 14, 2021

## Solution

Rajeswari answered on Jun 14 2021
60452 Assignment
Given is a readings of distance vs time in a graph plotted.
Here we are given 4 readings only as follows:
50 m = 28.90 sec or 1.73 m/sec
100 m = 61.35 sec or 1 .63 m/sec
200 m = 133.33 sec or 1.50 m/sec
400 m = 281.69 sec or 1.42 m/sec
Let us take time as x and distance travelled as y.
Then from the plot we can say that there is a linear...
SOLUTION.PDF