course Phy 121
2/6/10 at 1:00 p.m.
If an object increases velocity at a uniform rate from 6 m/s to 26 m/s in 12 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 6 m/s and whose velocity 12 seconds later is 26
m/s. Explain what the slope of the graph means and why, and also what the area means and why.
Acceleration = (26 m/s- 6 m/s)/12s = 20 m/s / 12 s = 1.67 m/s^2
Change in position = average velocity * change in clock time
average velocity = (6 m/s + 26 m/s)/2 = 16 m/s
Change in position = 16 m/s*12s=192 meters
slope = rise/run= (26-6)m/s / (12-0)s=20/12=1.67 m/s^2 This is the same as acceleration.
Area of a triangle is 1/2 (base)(height)= 1/2(12s)(20m/s)= 120 m. I'm uncertain how this relates.
The region beneath this graph is not a triangle, but a trapezoid.
Multiply the average of the two 'altitudes' by the 'width'. What does each mean, what is the product, and what does the product mean?
However, if I do take the change in position of 192 meters and divide by the area of 120 meters, I do get 1.6 which is the slope basically. Coincidence???
The two values are close but not identical; there is no relationship, but you asked a good question.
Good answers, except on the question about the area.
See if you can answer the questions I've posed.
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