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course Phy 201
9/18 1130
Sketch a velocity vs. clock time graph for an object whose initial velocity is 7 m/s and whose velocity 11 seconds later is 21 m/s. Explain what the slope of the graph means and why, and also what the area means and why.The graph will have a vertical axis of velocity with 2 distinct points of 7 and 21 in m/s. The horizontal axis will be from 0 to 11 secs. A straight line will connect from point (0,7) to (11,21).
What are we looking for?
Acceleration
displacement
acceleration=change in velocity/change in time
acceleration=(21m/sec-7m/sec)/(11sec-0sec)
acceleration=14/11m/sec^2=1.3m/sec^2
The slope in the graph represents the acceleration of the object.
displacement=avg velocity*time
avg velocity=(initial velocity+final velocity)/2
avg velocity=(7m/sec+21m/sec)/2=14m/s
displacement=14m/sec*11sec=154m
The area underneath the line of the graph represents the displacement of the object.
The area is equal to the displacement, provided the graph is a straight line, which is equivalent to uniform acceleration. You should include a qualifying statement in your solution, e.g., 'assuming a straight-line graph' or 'assuming uniform acceleration'.
If the graph isn't a straight line, having no other information about the graph, the best you can do is calculate the straight-line approximation, along with the qualifying statement.
&#Good responses. Let me know if you have questions. &#
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