Stopping Dist. Car @100km/h on Flat Surface

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Homework Help Overview

The discussion revolves around calculating the stopping distance of a car moving at 100 km/h on a flat surface, and how this distance changes when the car is on an incline of 10 degrees, both up and down. The original poster presents a method involving the coefficient of friction and vector analysis of forces acting on the car.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants explore the calculation of the coefficient of friction and the resulting acceleration of the car on an incline. Questions arise regarding the correctness of the equations used and the direction of forces when analyzing the problem.

Discussion Status

Some participants have provided feedback on the original poster's calculations, noting a dimensional inconsistency in the coefficient of friction formula. There is acknowledgment of the need to correctly identify the forces acting on the car when moving up the incline, indicating a productive exchange of ideas.

Contextual Notes

Participants are working under the assumption that the initial velocity and surface conditions remain constant, and they are focused on the implications of changing the incline angle on stopping distance.

inner08
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The minimal stopping distance for a car moving at an initial speed of module 100km/h is 60m on a flat surface. What is the stopping distance when the car is moving a) down at 10 degrees b) up at 10 degrees? We assume that the initial velocity and surface don't change.

I was thinking I could find the coefficient of friction of the flat surface (Uk = V^2/(2ag) which would give me Uk = 0.657. Then I would find the acceleration of the car by adding all the vectors together (mgsin(theta) - f = mA). The mass cancels out therefore giving me a = gsin(theta) - Ukgcos(theta) which in terms gives me a = -4.64. I then substitute the values in Vf^2 = Vi^2 + 2ad which gives me d = 83.3m. The answer to b would be mostly the same work switching up the sin/cos. Does that work?
 
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inner08 said:
The minimal stopping distance for a car moving at an initial speed of module 100km/h is 60m on a flat surface. What is the stopping distance when the car is moving a) down at 10 degrees b) up at 10 degrees? We assume that the initial velocity and surface don't change.

I was thinking I could find the coefficient of friction of the flat surface (Uk = V^2/(2ag) which would give me Uk = 0.657. Then I would find the acceleration of the car by adding all the vectors together (mgsin(theta) - f = mA). The mass cancels out therefore giving me a = gsin(theta) - Ukgcos(theta) which in terms gives me a = -4.64. I then substitute the values in Vf^2 = Vi^2 + 2ad which gives me d = 83.3m. The answer to b would be mostly the same work switching up the sin/cos. Does that work?
Your idea is correct. Your equation for Uk is dimensionally inconsistent. If you find the correct coefficient of friction, then you can find the stopping distance up and down the plane.
 
Oops, I meant Uk = V^2 /(2gs) not ag. Sorry. This, I think, fixes the inconsistency problem. Was the rest of my work ok?
 
inner08 said:
Oops, I meant Uk = V^2 /(2gs) not ag. Sorry. This, I think, fixes the inconsistency problem. Was the rest of my work ok?
Looks good for down the plane. Up the plane is not switching sin and cos. It is just getting the forces actin in the right direction.
 

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