Car Stopping Distance on a Wet Road

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SUMMARY

The discussion focuses on calculating the minimum stopping distance of a car traveling at 46.0 mi/h on a wet road with a coefficient of static friction of 0.101. The key equations used are x = v_ot - (1/2)at² and v = v_o - at, where v_o represents the initial velocity and a is the acceleration due to friction. The acceleration is derived from Newton's second law, resulting in a = 0.101g, with g being 9.81 m/s². The absence of the car's mass is noted, but it is not necessary for the calculation of stopping distance.

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A car is traveling at 46.0 mi/h on a horizontal highway.
If the coefficient of static friction between road and tires on a rainy day is 0.101, what is the minimum distance in which the car will stop?

Just confused on where to start because it seems like I need more info
 
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Jacob87411 said:
A car is traveling at 46.0 mi/h on a horizontal highway.
If the coefficient of static friction between road and tires on a rainy day is 0.101, what is the minimum distance in which the car will stop?

Just confused on where to start because it seems like I need more info
There is something missing. Don't you have a weight of the car ?

You should apply :
x = v_ot -a \frac{t^2}{2}
v = v_o -at

Where the v_o is initial velocity and a is the acceleration due to the friction. Beware of the sign !

Solve the second equation for t (the v is zero when the car has stopped) and substitute this t in the first equation to solve it for x.

marlon
 
Thats what I thought but there is no mass given
 
ohh yes, just apply Newton's second law in the vertical Y-direction :

ma_y = 0 = N -mg

normal force N equals N = mg

and friction force F = 0.101N = 0.101mg

The associated acceleration is F/m = 0.101g and g=9.81 m/s²

marlon
 

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