Question

The following are graphs from a full data collection run, using the wheel sensor of an ioLab device Wheel - Position (100 Hz) At: 0.66000 s με 0.247 m-0: 0.20 m 1.6 a: 0.16 ms s: 1.00 m/s 1.0 E 0.8 0.6 0.4 0.2 0.0 12 Time (s) 10 Rezero sensor Wheel Velocity (100 Hz) 0.66000 s μ: 0.989 m/s-0: 0.50 m/s 2.0 a: 0.65 m , s: 2.56 m/s* (r: 1.00) 1.2 0.8 S 0.6 0.4 0.2 0.0 10 12 Time (s) Wheel - Acceleration (100 Hz) At: 0.66000s με 2.554 m/s, . σ: 0.16 m/s* a: 1.69 m/s s: .0.27 mís (r 0.10) 25 20 15 N10 -5 -10 -15 .25 10 12The researcher then focuses on just a certain time period in the graphs, called the region of interest, highlighted in tan on the graphs above. A zoomed in version of the region of interest is shown below Wheel Position (100 Hz) t: 6.3804 s at: 0.66000 s 1.6 x 1.059 m 0.247 m-0.20m a:0.16 ms s: 1.00 m/s (r 0.95) 1.2 (6.9500 s, 0.5976 m) 0.6 6.4500 s, 0.045 m) 0.2 0.0 6.406.45 6.50 6.55 6.60 665 6.70 6.75 6.80 685 6.90 6.95 Rezero sensor Time (s) Wheel Velocity (100 Hz) At: 0.66000 s μ: 0.989 m/s-o: 0.50 m/s 2.0 a: 0.65 m s: 2.56 mis* (r, 1 .00) (6.9500 s, 1.74 m/s) 08 (6.4500 s, 0.46 m/s 0.6 0.4 0.2 0.0 0.2 6.406.456.50 6.55 6.60 6.65 6.706.756.80 6.85 6.90 6.95 Time (s) Wheel Acceleration (100 Hz) At: 0.66000 s μ: 2.554 m/s2-0: 0.16 m/s2 30 25 20 15 a: 1.69 m/s s:-0.27 m/s, (r®: O.10) i (6.4500 s, 2.43636 m/s) (6.9500 s, 2.34545 m/s2 .5 .15 -20 25 -30 6.40 6.45 6.50 6.55 6.60 6.65 6.706.75 6.80 6.85 6.90 6.95 Time (s)

The researcher chooses 2 data points within the region of interest to use in calculations. The first data point will be considered the initial point and the other data point will be considered the final point. Extract the numerical value for the following physical quantities from those 2 data pointS. vi

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Un tuding all tu a duti 12. tí= 6.45-00 s haph, the oispandin the inidial and tinal tam NDuw the velst crusponduing the initial & tpls upvote ?⭐if you like my answer

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