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course Phy 121
6/4 at 8:30 pm
If an object increases velocity at a uniform rate from 7 m/s to 21 m/s in 10 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 7 m/s and whose velocity 10 seconds later is 21 m/s. Explain what the slope of the graph means and why, and also what the area means and why.
a = change in velocity / change in time
a = (21 - 7) / (10 - 0) = 1.4 m/s
Distance = 1.4 m/s * 10 sec = 140 m
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You need to include units throughout your calculation. The numbers 21, 7, 10 and 0 all have units, and they determine the units of your result (which are not m/s).
displacement = ave velocity * time interval
1.4 m/s is not your average velocity (see previous note on units, and be sure you keep track of the meanings of your quantities).
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The x axis on this graph would represent time in seconds, beginning at 0 and going to 10 seconds. The y axis would represent velocity in meters per second, ranging from 0 to 21 m/s. The slope represents the relationship between two graph points and the average acceleration for the corresponding time interval. For an object with an initial velocity of 7 m/s, the point would be at (0,7). The second point would be at (10, 21) as time passes and it accelerates. The run here is the change in time, while the rise is the change in velocity. The area beneath the graph represents the change in position, which I solved for by multiplying acceleration by time.
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Change in position is not acceleration * time interval.
You did in fact get the correct result for the displacement, but that isn't how it's calculated. Also 1.4 * 10 is 14, not 140.
I think you'll find this easy and quick to correct.
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