A car stopping on a slippery surface on an uphill

In summary, the equation for the direct deceleration due to gravity is mg(sin angle of elevation)=-.5at^2+V_0t.
  • #36
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.
 
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  • #37
Heidi Henkel said:
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).
Heidi Henkel said:
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.
 
  • #38
There are 2 different definitions of grade, with people measuring it differently? I want to know about this.
 
  • #39
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.
 
  • #40
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 ?
 
  • #41
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.
 
  • #42
I posted to the homework forum and changed the problem then, to 8% grade etc. Look at the problem posted in homework template format.
 

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