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course PHY 201
3/19
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.
aAve = (27 - 8)/ 13 = 1.5 cm/sec
The slope is the aAve and the area is the change in position. If the velocity is in the y axis and the clock time in the x axis then a change in velocity is the same as the rise, similarily a change in time is the same as the run. So rise/run equals slope which equals change in velocity/ change in time, acceleration.
Finding the average velocity (as used in finding change in position) is the same as finding the midpoint of the line (0,8; 13,27) and using that to find the area of the rectangle you created.
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(27 - 8)/ 13 = 1.5, not 1.5 cm/sec.
(27 cm - 8 cm) / (13 s) = 1.5 cm/s, but 27 cm and 8 cm are not positions associated with this problem.
27 cm/s and 8 cm/s are velocities associated with this poblem, and
(27 cm/s - 8 cm/s )/ (13 s) = 1.5 cm/sec^2
is the rate of change of velocity.
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