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Calculating v from v0, frictional coefficients, and angle

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!
 

TSny

Homework Helper
Gold Member
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Which direction are you taking to be the positive direction along the slope? Do both of your signs for the initial velocity and acceleration (in the x equation) agree with your choice of positive direction?

[EDIT: Never mind, I saw a = 3.33 (positive) but you took care of the sign by inserting a negative in the x formula. Sorry]
 

TSny

Homework Helper
Gold Member
12,046
2,619
Are you sure you get negative values for t when you solve the quadratic equation? I get positive values.
 
Oh--you're right! Recalculated with an online quad formula calculator and it worked. Thank you for your help!
 

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