Stopping distance considering coefficient of frictionby Aubiefan Tags: coefficient, distance, friction, stopping 

#1
Sep1706, 09:21 AM

P: 16

I am having a lot of trouble with this problem:
A car is traveling at 45.0 km/h on a flat highway. If the coefficient of friction between road and tires on a rainy day is 0.100, what is the minimum distance in which the car will stop? I know it uses the delta X kinematics equation, so I should use friction to figure out the acceleration, but we haven't covered this material in class and I am at a loss for how to proceed. Also, will I be considering four systems, one for each tire in contact with the road? Thanks, any tips are appreciated! 



#2
Sep1706, 11:31 AM

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P: 39,683

What equation do you use to relate the frictional force F to the coefficient of friction mu and the normal force (weight) N? Even though N points down, which direction does the force F of friction point? And no, you don't need to do all 4 wheels, because you would just get 1/4 the force at each, which adds up to 1 for all of them. Show us your work and we can help guide you if you need help.




#3
Sep1706, 01:03 PM

P: 16

The frictional force (F) is mu times the normal force, but I don't know what the mass or weight of the car is, that iis one main thing that is confusing me. I am given the coefficient, so I have F= (0.1)n, which can be rearranged to 0.1=n/F, or 0.1=applied force/frictional force. So I know that frictional force is ten times the applied force, but I am still not sure what equation to use to try and get acceleration from that. Any tips on the best equation to use here?
Thanks for clearing up the issue about the four wheels, I appreciate it! 



#4
Sep1706, 01:06 PM

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P: 39,683

Stopping distance considering coefficient of friction 



#5
Sep1706, 03:08 PM

P: 16

I know that F=ma, so if F equals ma, so I would think that ma=0.1n, meaning acceleration is 0.1n/m. I also know that n=mg, so ma=0.1(mg), now I see where mass cancels out, thanks for the hint.
So I set a=(0.1)g, and got a=.98. I plugged that into V^2=Vo^2 + a(delta X), with 12.5 m/s^2 as Vo (my starting speed, converted from km/hr) and V=0. I solved for delta X, and got 76.56. When I typed it into my Webassign answer sheet, it told me it was incorrect but within 10% of the right answer. Any idea where I have gone wrong? thanks for your patience! 



#6
Sep1706, 03:17 PM

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P: 39,683

> V^2=Vo^2 + a(delta X), with 12.5 m/s^2 as Vo
First, a typo  the units of velocity are m/s. Second, the full equation for motion is [tex]d = d_0 + v_0 t + \frac{a t^2}{2}[/tex] [tex]v = v_0 + a t [/tex] Use the 2nd equation to find the time it takes to stop, then plug that into the first one to get the distance. What answers do you get? EDIT  fixed a couple LaTex typos 



#7
Sep1706, 10:21 PM

P: 16

I finally got it! I solved it using your method, and then tried the first method again and got the same answer, I guess I had been punching something into the calculator wrong. My final answer was 79.7 m.
Thank you so much for your help and your patience, it is EXTREMELY appreciated!!! 


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