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

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The minimal stopping distance for a car traveling at 100 km/h on a flat surface is 60 meters. To calculate stopping distances on a slope, the coefficient of friction must be correctly determined, with the formula Uk = V^2 /(2gs). The acceleration of the car is influenced by the angle of the slope, requiring careful consideration of the forces involved. For a downward slope at 10 degrees, the stopping distance is calculated to be 83.3 meters, while the approach for an upward slope involves correctly identifying the direction of forces rather than merely switching sine and cosine. Accurate calculations are essential for determining stopping distances on inclined surfaces.
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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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Thread 'Variable mass system : water sprayed into a moving container'

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