#$&*
Phy 121
Your 'cq_1_07.2' report has been received. Scroll down through the document to see any comments I might have inserted, and my final comment at the end.
** CQ_1_07.2_labelMessages **
An automobile rolls 10 meters down a constant incline with slope .05 in 8 seconds, starting from rest. The same automobile requires 5 seconds to roll the same distance down an incline with slope .10, again starting from rest.
• At what average rate is the automobile's acceleration changing with respect to the slope of the incline?
answer/question/discussion: ->->->->->->->->->->->-> :
#$&*
I know:
`ds = 10m `ds = 10m
`dt = 8s `dt = 5s
slope = .05 slope = 0.1
v0 = 0m/s v0 = 0m/s
To solve for vf
`ds = (v0 + vf) / 2 * `dt
(`ds / `dt) * 2 - v0 = vf
(10m/ 8s) * 2 - 0 = 2.5m/s = vf (10m / 5s) * 2 - 0 = 4m/s = vf2
To solve for a
vf = v0 + a(`dt)
(vf / `dt) - v0 = a
(2.5m/s / 8s) - 0m/s = 0.3125m/s^2 = a (4m/s / 5s) - 0m/s = 0.8m/s^2
So, `da = 0.8m/s^2 - 0.3125m/s^2 = 0.4875m/s^2
and `d slope = 0.1 - 0.05 = 0.05
so, 0.4875m/s^2 / 0.05 = 9.75
** **
30 mins
** **
&#Your work looks very good. Let me know if you have any questions. &#