A car stopping on a slippery surface on an uphill

[Heidi Henkel]

Same event on flat surface: V_f^2 = 11.176^2 + 2 (-3.2) (5.8)
v_f = 9.37m/s = 20.96 mph
so the hill slows the car down more than the loss of F_n impairs the braking.

 

[Heidi Henkel]

a=2.2m/s / change in time
d=-.5at^2+v_it
5.8=-.5 (2.2/t) t^2 + 11.176t
5.8=10.076t
0.5756 seconds to hitting deer when going uphill, compared to 0.56 seconds to hitting deer on level ground

 

[haruspex]

78.779 = v_f^2
v_f=8.876 m/s
19.856mph

Haven't checked it in detail, but it looks right.
You can get h a little more simply. You are given that it is an 8% grade, which means the rise is 0.08 d.

so the hill slows the car down more than the loss of F_n impairs the braking.

Not sure how you deduce that. The braking gives a deceleration $\mu g$, or about 0.3 g, whereas the slope gives only 8% of g.

 

[Heidi Henkel]

h is 0.08 times the horizontal, not 0.08 times the hypotenuse. So it is not 0.08d. d is the hypotenuse. You have to find the horizontal component of d.

I deduce my conclusion by calculating V_f for a slope and then for a flat, all variables the same except the slope. I wrote it on this forum, above. Did I do it wrong?

 

[haruspex]

h is 0.08 times the horizontal, not 0.08 times the hypotenuse. So it is not 0.08d. d is the hypotenuse. You have to find the horizontal component of d.

I believe that slope grades on roads are quoted as height/slope distance, i.e. the sine of the angle. On such a shallow grade it will make hardly any difference anyway.

I deduce my conclusion by calculating V_f for a slope and then for a flat, all variables the same except the slope. I wrote it on this forum, above. Did I do it wrong?

The square root process to find velocity might mislead you here. To be certain, you need to compare the reduction in speed by braking only with the reduction in speed by coasting up the hill. But it is pretty clear that a braking coefficient of .32 is going to provide about four times the deceleration of an 8% grade.

 

[Heidi Henkel]

Grades are slopes. up/across times 100 (which just makes it a %) So yes you can use trig, or you can introduce some error by using 8% of distance.

I think you misunderstood my conclusion. I think we are not disagreeing.

Car braking for 5.8m going up 8% grade slows to 19mph and car braking for 5.8m on flat slows to 20mph, so the uphill outweighs the change in friction. The reduced friction because of the hill is more than compensated for by the elevation gain. The uphill car slows more, in spite of a smaller friction force. However, at 8% grade the friction is still a larger part of what slows the car, than the grade.

 

[haruspex]

Grades are slopes. up/across times 100

Ok, looks like a difference between the "1 in N" system I grew up with in England (the sine of the angle) and the current international standard (the tangent).

I think you misunderstood my conclusion. I think we are not disagreeing.

Ok, I see, you are referring to the slight reduction in braking power caused by the lower normal force on the slope.

 

[Heidi Henkel]

There are 2 different definitions of grade, with people measuring it differently? I want to know about this.

 

[Heidi Henkel]

The tangent of the angle yes that is how I have been understanding grade. If that is the current international standard, I will go with that.

 

[Rerry]

Your question has a lot of variables, such as surface friction, regular tires/ studded, vehicle weight, percent grade of roadway and driver reaction. For each variable, you would use a different equation, resulting in a different stopping distance . Are you allowing for these variables in your answer or is your answer Plus or minus these variables ?

 

[Heidi Henkel]

Most of the answers to your question are in the stated problem and the mass cancels out. My calculation begins when brakes are applied, so driver reaction is not part of the calculation.

 

[Heidi Henkel]

I posted to the homework forum and changed the problem then, to 8% grade etc. Look at the problem posted in homework template format.