course phy 201
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?(27m/s-8m/s)/ 13 seconds= 1.46m/sec^2
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.
Rise/Run= 27m/s-8m/s)/13sec= 1.46 m/sec^2 = slope
Between the x coordinates of 8 and 27, there is a y stretch of 13 seconds, the slope increases 1.46m/sec^2. The graph is a straight line which increases at a constant rate.
Good solution, but see the linkg for a clarification of the explanation:
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Solution
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