course PHY201
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 = 10m ‘ds2 = 10m
‘dt1 = 8s ‘dt2 = 5s
Vavg = ‘ds/’dt = 10/8 = 1.25m/s vavg = 10/5 = 2m/s
Vf = vavg*2-0 = 1.25*2 = 2.5m/s vf = 2*2 = 4m/s
‘dv = vf-v0 = 2.5-0 = 2.5m/s ‘dv = 4-0 = 4m/s
A= ‘dv/’dt = 2.5/8 = .3125m/s/s a= 4/5 = 0.8m/s/s
‘da = .8 - .3125 = .4875m/s/s
‘dslope = .10-.5 = -0.4
Rate = ‘da /’dslope = .4875/-.4 = -1.21875m/s/s
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Very good work.