#$&*
PHY 121
Your 'cq_1_07.1' report has been received. Scroll down through the document to see any comments I might have inserted, and my final comment at the end.
** **
________________________________________
A ball falls freely from rest at a height of 2 meters. Observations indicate that the ball reaches the ground in .64 seconds.
Based on this information what is its acceleration?
answer/question/discussion: ->->->->->->->->->->->-> :
vAve = 2 m / .64 s = 3.125 m/s
3.125 m/s = ( 0m/s + vf) / 2
Vf = 3.125 m/s * 2 + 0 m/s = 6.250 m/s
aAve = (6.250 m/s - 0 m/s) / .64 s = 9.8 m/s^2
#$&*
Is this consistent with an observation which concludes that a ball dropped from a height of 5 meters reaches the ground in 1.05 seconds?
answer/question/discussion: ->->->->->->->->->->->-> :
vAve = 5m / 1.05 s = 4.8 m/s
vf = 4.8 m/s * 2 + 0 m/s = 9.6 m/s
aAve = (9.6 m/s - 0 m/s) / 1.05 s = 9.1 m/s^2
No, this one accelerated a little slower.
#$&*
Are these observations consistent with the accepted value of the acceleration of gravity, which is 9.8 m / s^2?
answer/question/discussion: ->->->->->->->->->->->-> :
The first one is, and the second one is slower by .7 m/s^2
#$&*
*#&!
&#This looks very good. Let me know if you have any questions. &#
PHY 121
Your 'cq_1_07.2' report has been received. Scroll down through the document to see any comments I might have inserted, and my final comment at the end.
** **
________________________________________
An automobile rolls 10 meters down a constant incline with slope .05 in 8 seconds, starting from rest. The same automobile requires 5 seconds to roll the same distance down an incline with slope .10, again starting from rest.
At what average rate is the automobile's acceleration changing with respect to the slope of the incline?
answer/question/discussion: ->->->->->->->->->->->-> :
.05 slope
vAve = 10 m / 8 s = 1.25 m/s
vf = 1.25 m/s * 2 + 0 m/s = 2.5 m/s
aAve = (2.5 m/s - 0 m/s) / 8 s = .3 m/s^2
.10 slope
vAve = 10 m / 5 s = 2 m/s
vf = 2 m/s * 2 + 0 m/s = 4 m/s
aAve = (4 m/s - 2 m/s) / 5 s = .4 m/s^2
@& The initial velocity on this incline is also zero. The acceleration is .8 m/s^2.*@
aAve of .10 slope v .05 slope
aAve = ( .4 m/s^2 - .3 m/s^2) / .05 = 2 Dont think this is right
@& You wouldn't call this aAve. This is the rate of change of aAve with respect to ramp slope.
The units are correct, and if you had used .8 m/s^2 for the acceleration on the second ramp you would have the correct result.*@
but for every .01 change in slope, the aAve increases between the two by .02 m/s^2
#$&*
@& *@
@& Very good, with just one real error. Check my notes.*@