course phy 201
If an object increases velocity at a uniform rate from 6 m/s to 26 m/s in 10 seconds, what is its acceleration and how far does it travel?Average acceleration = (v2 - v1)/t
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= (26 m/s - 6 m/s) / 10s
= (20 m/s) / 10s
= 2 m/s^2
Using the average acceleration, I calculated the object to travel 100m.
distance = ( (t^2)*acceleration ) / 2
= ( (10s^2)*2m/s^2 ) / 2
= 100 m
Sketch a velocity vs. clock time graph for an object whose initial velocity is 6 m/s and whose velocity 10 seconds later is 26 m/s. Explain what the slope of the graph means and why, and also what the area means and why.
The graph has a positive slope, increasing at a constant rate. The slope of the graph illustrates the object's acceleration. It uses the same basic formula for a v vs. t graph.
m = (difference in y values) / (difference in x values)
m = (26m/s - 6m/s) / (10s - 0s)
m = (20m/s)/10s
m = 2 m/s^2
The area under a velocity vs. time graph represents the object's displacement.
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Good. Let me know if you have questions.