#$&*
Phy 121
Your 'cq_1_07.2' report has been received. Scroll down through the document to see any comments I might have inserted, and my final comment at the end.
** **
An automobile rolls 10 meters down a constant incline with slope .05 in 8 seconds, starting from rest. The same automobile requires 5 seconds to roll the same distance down an incline with slope .10, again starting from rest.
At what average rate is the automobile's acceleration changing with respect to the slope of the incline?
answer/question/discussion: ->->->->->->->->->->->-> :
'ds1 = 10 m, 'dt1 = 8 sec, slope1 = .05
'ds2 = 10 m, 'dt2 = 5 sec, slope2 = .10
Trial 1: vAve = 10 m / 8 sec = 1.25 m/s. vf = v0 + vAve * 2, so vf = 2.5 m/s and v0 = 0 m/s. a = 2.5 m/s / 8 sec = 0.3125 m/s^2
Trial 2: vAve = 10 m / 5 sec = 2 m/s. vf = v0 + vAve * 2, so vf = 4 m/s and v0 = 0 m/s. a = 4 m/s / 5 sec = 0.8 m/s^2
The rate of the automobile's acceleration changes at (0.8 m/s^2 - 0.3125 m/s^2) / (.05 - .1) = -9.75 m/s^2 per slope change (? on units)
I do not like my answer here because I implied that v0 = 0 m/s and it also shows that the acceleration is decreasing when it actually increased.
0.8 m/s^2 is associated with slope .1. You used a different order of subtraction in your numerator than your denominator. If you correct this you have an excellent result.
I could use the formula a = g * slope.
Trial 1: a = 9.8 m/s^2 * .05 = .49 m/s^2
Trial 2: a = 9.8 m/s^2 * .1 = .98 m/s^2
This shows that when the slope doubles the acceleration doubles.
#$&*
** **
10 min
** **
No need for revision; your only error is minor and obvious. Check the given solution to confirm this:
&#See any notes I might have inserted into your document. If there are no notes, this does not mean that your solution is completely correct.
Then please compare your solutions with the expanded discussion at the link
Solution
Self-critique your solutions, if this is necessary, according to the usual criteria. Insert any revisions, questions, etc. into a copy of this posted document. Mark any insertions with &&&& so they can be easily identified.If your solution is completely consistent with the given solution, you need do nothing further with this problem. &#