Calculating Minimum Car Stopping Distance on a Rainy Day

[conno719]

A car is traveling down a flat highway at 11.39 m/s(squared), if the coefficient of friction is .100 (its a rainy day) what is the minimum distance in which the car will stop?

Im not sure whether to start with an equation of motion or use F(x)= ma(x).
Just a little confused. I am not sure if the weight of the car is needed, but it is not given...

 

[stewartcs]

Since the car is sliding the acceleration is assumed to be constant. This means you can use the constant acceleration equations of motion.

v = v0 + at

Relating the stopping force (opposing friction) to Newton's second law gives us

-fk = ma

Solving for a gives

a = -fk/m

Remembering that fk = uk*Fn we can substitute that into the previous equation and get

a = -uk*Fn / m

The normal force Fn = the cars mass time g, which gives

a = -uk*mg / m which reduces to a = -uk*g

As you can see the mass of the car is irrelevant.

Now solve the constant acceleration equation that was first given for t using a that was just found. Remeber that the final velocity will be zero since the problem wants to know how long it will take to stop the car.

 

[conno719]

thank you for the help

 

[stewartcs]

Sorry, I just noticed that the question was asking for the distance and not time. So you'll have to use the time you just found to find the displacement using one of the other constant acceleration equations of motion.

x = v0*t + (1/2)*a*t^2 should do it.

BTW I assumed that 11.39 was the velocity in m/s and not the acceleration in m/s^2 (you have 11.39 m/s(squared) written).

If this is not the case then the answer will be different.