- #1
KMjuniormint5
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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 = 25.5 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 = 18.0 m/s.
(a) With what speed did car A hit car B if the coefficient of kinetic friction was 0.60 (dry road surface)?
how I went about the problem:
a = acceleration only in x direction
for car A:
Fnet = mAa --> WA2 - Fk = (mA)a --> (mA)(g)(sin12.0) - [tex]\mu_k[/tex] (mA)(g) = (mA)a
now i divided everything by (mB) to get final equation of:
(g)(sin12.0) - [tex]\mu_k[/tex] (g) = a
now since constant acceleration I plugged a into the equation of
V^2 = 2a(delta x) + Vo^2 to get --> V^2 = 2((g)(sin12.0) - [tex]\mu_k[/tex] (g))(x) =Vo^2
plugged in everything I know:
v^2 = 2 (2(9.8sin12.0 - ((0.6)(9.8)))(25.5) - 18^2
then take the square root to get v = 11.3152m/s
why is this not right?
(a) With what speed did car A hit car B if the coefficient of kinetic friction was 0.60 (dry road surface)?
how I went about the problem:
a = acceleration only in x direction
for car A:
Fnet = mAa --> WA2 - Fk = (mA)a --> (mA)(g)(sin12.0) - [tex]\mu_k[/tex] (mA)(g) = (mA)a
now i divided everything by (mB) to get final equation of:
(g)(sin12.0) - [tex]\mu_k[/tex] (g) = a
now since constant acceleration I plugged a into the equation of
V^2 = 2a(delta x) + Vo^2 to get --> V^2 = 2((g)(sin12.0) - [tex]\mu_k[/tex] (g))(x) =Vo^2
plugged in everything I know:
v^2 = 2 (2(9.8sin12.0 - ((0.6)(9.8)))(25.5) - 18^2
then take the square root to get v = 11.3152m/s
why is this not right?