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course Phy121
9/19/2009 around 4:20pm
If an object increases velocity at a uniform rate from 8 m/s to 27 m/s in 13 seconds, what is its acceleration and how far does it travel?Sketch a velocity vs. clock time graph for an object whose initial velocity is 8 m/s and whose velocity 13 seconds later is 27 m/s. Explain what the slope of the graph means and why, and also what the area means and why.
Acceleration would be :( 27m/s-8m/s)/13s = (19m/s) /13s =1.5m/s^2
The distance traveled would be: 27m/s -8m/s =9m/s
That's the change in velocity, not the distance.
On a graph the rise between two points represents the change in velocity and the run represents the change in position. The slope shows the average acceleration for the corresponding time interval. The area beneath the graphs represents change in position and the width of a region beneath the graph represents change in clock time.
Not bad, but the one you missed contains a very important distinction, so you should submit a revision to be sure you understand.
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