course Phy121
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Over a period of 10 seconds, an object increases its velocity at a uniform rate from 3 m/s to 27 m/s * What is its acceleration and how far does it
* Graph velocity vs. clock time for this object and explain what the slope of the graph means and why, and also what the area means and why.
The acceleration is the change in velocity, 24m/s divided by the change in time, which is 10 seconds.
`dv/`dt = 2.4 m/s^2
The slop of this graph, the rise/run, is the acceleration of the object, which was given above. The area under this graph, represents the change in position during the 10 seconds. For this example, I divided the graph into 5 seconds intervals, and the results were a `ds of 120 meters.
The first trapezoid had an altitude of 6m/s over a period of 5 seconds. Since the 6m/s is the midpoint between the starting and ending velocity for that time interval, and since it is acclerating at a uniform rate, this 6m/s also represents the vAve for that interval. So, since vAve = `ds/`dt, we can figure out `ds for this time interval. In this case, `ds = vAve * `dt, or 6m/s * 5 s = 30 m. I repeated this for the 2nd time interval between 5 and 10 seconds and the `ds was 90m.
So, 90 m + 30 m = 120 meters for the `ds over the 10 second period.
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Excellent work and excellent explanation. This reflects very good understanding of these concepts.