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Phy 231
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** 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: ->->->->->->->->->->->-> :
vAve = 10m / 8 sec = 1.25 m/s = (vf +v0) /2
vAve*2 = vf = 2.50 m/s
'dv = vf - vo = 2.50 m/s
a = 'dv / 'dt = 2.50 m/s / 8 sec = .3125 m /s^2
a / slope = accel with respect to slope = .3125 m/s^2 / .05 slope = 6.25 m/s^2 / 1 unit of slope
vAve 2 = 10 m / 5 sec = 2 m/s = (vf +vo) / 2
Vave*2 = 4.00 m/s
'dv = vf - vo = 4.00 m/s
a = 'dv / 'dt = 4.00 m/s / 5 sec = 0.800 m/s^2
a / slope = accel with respect to slope = 0.800 m/s^2 / .10 = 8 m/s^2 / 1 unit of slope
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.800 m/s^2 is one of your accelerations. It is not a difference between accelerations.
An average rate of change is defined to be the change in one quantity divided by the change in the other.
You have divided one quantity by another, not a change in one quantity by a change in another.
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