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^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=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]

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]

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!

 

[Nickie]

I would first like to clarify that these calculations are based on simplified assumptions and may not accurately reflect the real-world scenario of the accident.

Now, let's take a closer look at the equations and variables used in the attempt at the solution:

- Fn=mgcos12: This equation assumes that the normal force (Fn) acting on the car is equal to the weight (mg) of the car multiplied by the cosine of the angle of the hill (12 degrees). However, in reality, the normal force would also be affected by the slope of the road and the weight distribution of the car. Therefore, this equation may not accurately represent the normal force in this scenario.

- fk=μkFn: This equation assumes that the kinetic friction force (fk) is equal to the coefficient of kinetic friction (μk) multiplied by the normal force (Fn). Again, this may not accurately reflect the real-world scenario as the friction force would also be affected by other factors such as the surface of the road and the condition of the tires.

- x-xo=vot-(1/2)at^2: This is the equation for displacement (x) as a function of initial position (xo), initial velocity (vo), acceleration (a), and time (t). However, in this scenario, we are interested in finding the final velocity (v) at the time of impact, not the displacement. Therefore, this equation may not be the most appropriate one to use.

Based on these observations, I would suggest using a different approach to solve this problem. Instead of trying to find the final velocity at the time of impact, we can use the conservation of energy principle to calculate the speed of car A when it hit car B. This approach would take into account the change in potential energy (due to the slope of the hill) and the work done by friction (which would be equal to the change in kinetic energy).

Using this approach, the final velocity of car A can be calculated as:

v=sqrt(2gh+μkd)

Where:
- g is the acceleration due to gravity (9.8 m/s^2)
- h is the height of the hill (d*sin12)
- μk is the coefficient of kinetic friction
- d is the separation distance between the two cars (23 m)

Plugging in the values for the two coefficients of kinetic friction given in the problem, we get:

(a) v=sqrt(2