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course Phy 231
2/17 10
If an object increases velocity at a uniform rate from 8 m/s to 22 m/s in 5 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 5 seconds later is 22 m/s. Explain what the slope of the graph means and why, and also what the area means and why.
a = (vf-v0)/'dt = ( 22 - 8 ) m/s / 5 s = 2.8 m/s
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Good, except that the units you get when you divide m/s by s will not be m/s.
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v_ave = (Vf + vo) / 2 = ( 22 + 8) m/s /2 = 15 m/s
'ds = v_ave * 'dt = 15 m/s * 5s = 75 meters
a = 2.8 m/s, 'ds = 75 meters
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m/s is not a unit of acceleration. Otherwise OK.
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The slope of the graph is the velocity with respect to time and is equal to the acceleration.
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Terminology note:
The slope of the graph is not the velocity with respect to time.\
The slope of the graph is the average rate of change of velocity with respect to time.
If you don't use all the words, it appears that you might not fully understand the terminology, the concept and the definition.
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rise / run = slope, rise / run = change in velocity / change in time. so the rate of change is = (m/s)/s = m/s^2
The area under the graph is equal to the total distance covered by the object.
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This is good, but there are a couple of issues you'll want to address.
No need for a revision unless you're unsure of anything, but be sure to see my notes.
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