# Calculating v from v0, frictional coefficients, and angle

#### rockchalk1312

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

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!

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#### TSny

Homework Helper
Gold Member
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
Are you sure you get negative values for t when you solve the quadratic equation? I get positive values.

#### rockchalk1312

Oh--you're right! Recalculated with an online quad formula calculator and it worked. Thank you for your help!

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