Car Stopping Distance on a Wet Road

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Homework Help Overview

The discussion revolves around calculating the stopping distance of a car traveling at 46.0 mi/h on a wet road, where the coefficient of static friction is given as 0.101. Participants express confusion regarding the necessary information and parameters needed for the calculation.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants question the absence of the car's weight and mass, which are considered essential for applying the relevant equations. There are attempts to apply kinematic equations and Newton's second law to derive the stopping distance.

Discussion Status

The discussion is ongoing, with participants exploring different aspects of the problem. Some guidance has been offered regarding the application of equations, but there is no consensus on how to proceed due to missing information.

Contextual Notes

Participants note the lack of mass and weight information, which are critical for determining the stopping distance accurately. The discussion reflects uncertainty about the necessary parameters for the calculations.

Jacob87411
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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 :
[tex]x = v_ot -a \frac{t^2}{2}[/tex]
[tex]v = v_o -at[/tex]

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 :

[tex]ma_y = 0 = N -mg[/tex]

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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