#$&*
course Phy 231
2-5-11 about 7:30pm
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= (27m/sec – 8m/sec)/13sec= 1.46m/sec/sec
vAve= (27m/sec + 8m/sec)/2= 17.5m/sec
ds= 17.5m/sec * 13sec= 227.5m
rise/run= (19m/sec)/13sec= 1.46m/sec/sec= slope
The slope is = to the acceleration of the object.
The area of the trapezoid is = to the displacement of the object which is 227.5m.
The altitudes of the trapezoid are 8m/sec and 27m/sec, and it has a base of 13sec. The area of the trapezoid is (27m/sec + 8m/sec)/2 * 13= 227.5m
@& You haven't really explained why the slope gives the average acceleration, or the area gives the displacement.
Be sure you can give good explanations. If you aren't sure, submit your best explanations and ask me for feedback.*@