rockchalk1312
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You testify as an expert witness in a case involving an accident in which car A slid into the rear of car B, which was stopped at a red light along a road headed down a hill. You find that the slope of the hill is θ = 12.0°, that the cars were separated by distance d = 23.0 m when the driver of car A put the car into a slide (it lacked any automatic anti-brake-lock system), and that the speed of car A at the onset of braking was v0 = 20.0 m/s. With what speed did car A hit car B if the coefficient of kinetic friction was (a) 0.560 (dry road surface) and (b) 0.190 (road surface covered with wet leaves)?equations:
Fn=mg
Fk=μkFn
x-xo=volt-(1/2)at^2attempt at solution:
Fn=mgcos12
=(9.8)(cos12)(m)
=9.59m
fk=μkFn
fk-mgsin12=ma
μkgcos12-gsin12=a
(.560)(9.8)(cos12)-(9.8)(sin12)=a
a=3.33
x-xo=volt-(1/2)at^2
23=20t-(1/2)(3.33)t^2
0=-1.665t^2+20t-23
quadratic formula gave -1.288, -10.72
v=vo+at
v=20+(-3.33)(1.05)
v=16.5035
obviously can't have -t so I just plugged in +1.05 into that last equation, but I'm just trying to figure out what I did wrong because this was the wrong answer. Thank you!
Fn=mg
Fk=μkFn
x-xo=volt-(1/2)at^2attempt at solution:
Fn=mgcos12
=(9.8)(cos12)(m)
=9.59m
fk=μkFn
fk-mgsin12=ma
μkgcos12-gsin12=a
(.560)(9.8)(cos12)-(9.8)(sin12)=a
a=3.33
x-xo=volt-(1/2)at^2
23=20t-(1/2)(3.33)t^2
0=-1.665t^2+20t-23
quadratic formula gave -1.288, -10.72
v=vo+at
v=20+(-3.33)(1.05)
v=16.5035
obviously can't have -t so I just plugged in +1.05 into that last equation, but I'm just trying to figure out what I did wrong because this was the wrong answer. Thank you!