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course PHY 241
225pm 10/22/2011
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?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.
v_1 = 8 m/s
v_2 = 27 m/s
t_1 = 0 s
t_2 = 13 s
acceleration =
= 'dv/'dt
= (v_2-v_1) / (t_2-t_1)
= (19 m/s) / 13s
= 1.5 m/s^2
distance traveled =
= (v_2 + v_1) / 2 * 13 s
= 17.5 m/s * 13 s
= 227.5 m
I sketched a graph of velocity vs clock time with points (0,8) and (13,27) and the drew a line segment between those two points. I create a right triangle with a base of 13 and a height of 19. The slope of the graph is the acceleration of the object. Any point on the graph has a slope of 19/13 =approx 1.46 m/s^2. Then I found the Area of the triangle to be (b*h)/2 = 13*19/2 = 123 m. Then I found the area of beneath the triangle, which is a rectangle with side of 13 and height of 8, so the are is bh = 104 m and I added the two areas together. 104m + 123.5 m = 227.5 m which is the 'ds or change in position of the object.
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Self-critique (if necessary):
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Self-critique rating:
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@& If you regard the region beneath the graph as a trapezoid then its 'graph altitudes' represent velocities and the average of these altitudes represents the average velocity.
Then the area represents average velocity * `dt, which is `ds.
This insight gives meaning to the integral, which would also give us the area beneath the graph, but which is unnecessary when the graph is a straight line.*@