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=vot-(1/2)at^2 attempt 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=vot-(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!