Braking Down an Incline

  • Thread starter blueray4
  • Start date
  • #1
9
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I've been trying to figure this one out for two days now. No Dice. If you feel like working the whole thing out, by all means, cause everyone who tried to help me so far was wrong.

Homework Statement



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 (Fig. 6-27). You find that the slope of the hill is = 12.0°, that the cars were separated by distance d = 26.0 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 = 17.5 m/s.

(Basically, Car A was traveling down an incline, Car B is stopped the entire time. It started braking when it was 26.0 meters away w/ initial velocity of 17.5 m/s)

What is final velocity before impact when coefficient of friction is .6?
What is final..........................................................friction is .1?

Homework Equations



IDK, ive been trying to use V^2= Vo^2 + 2a (delta x)


The Attempt at a Solution



damn, i've had so many attempts, i have no clue which one to put up.
 
Last edited:

Answers and Replies

  • #2
1,860
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Are the answers 19.5m/s and 20.2m/s?
 
  • #3
9
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i don't have the answers. but, they don't even seem reasonable, because they are more than the initial.
 
  • #4
1,860
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Why is that unreasonable? The force of gravity was greater than the force of friction, and so there would be an acceleration downhill.
 
  • #5
9
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i'm confused. i guess i assumed braking would decelerate. im so sick and tired of this question.
 
  • #6
1,860
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I can see why you are confused, but you should remember that the car is sliding down the hill. It's the same thing as a block sliding down a ramp.

What you are probably thinking is the car's deceleration when the tires are in contact with the road, and thus having a much greater coefficient of friction because it would be static friction.

Which answers did you come up with?
 
  • #7
960
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looking at the problem along the incline,
you have stopping power of:
mu*N*cos(12)=mu*mg*cos(12)

accelerating power of mg*sin(12)

net acceleration of g(mu(sin12))-cos(12))
=9.8(.6(.98))-.21)=-.38g

So plugging into your favorite kinematic eqn, along the incline, you have an initial velocity of 17.5, a retarding acceleration of .38g, and a distance of 26m.
Is ths what you had?
 
  • #8
9
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no nothing like those numbers.
 
  • #9
9
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but if the acceleration is retarding, doesnt that mean it is slowing?
 
  • #10
960
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yea, but maybe not enuf--crash!

If you look up your kinematics likely find something like Vf^2-Vi^2=2ax

here, Vi=17.5, x=26m. You have 3/8 negative gee. Assume you had a full gee of stopping force, you'd have 17.5^2/(2*9.8)=15.625 meters to stop. We have a longer distance but lot less stopping force.
 
  • #11
1,860
0
Ah crap, I might have typed the numbers into my calculator wrong. Algebraic answers are much preferred by me. Anyway, you should get that

[tex] a = g sin \theta - \mu_k g cos \theta [/tex]

then you can use kinematics from there.
 

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