Q35E - University Physics with Modern Physics

Question

(a) If the coefficient of kinetic friction between tires and dry pavement is 0.80 , what is the shortest distance in which you can stop a car by locking the brakes when the car is traveling at 28.7 m/s (about 65mi/h )? (b) On wet pavement the coefficient of kinetic friction may be only 0.25 . How fast should you drive on wet pavement to be able to stop in the same distance as in part (a)? (Note: Locking the brakes is not the safest way to stop.)

Step-by-Step Solution

Answer

(a) 52.53m

(b) 16.04 m/s

Step 1: Given data

Coefficient of kinetic friction between tires and dry pavement is ${\mu }_k=0.8$ 

Initial speed  u = 28.7 m/s

Final velocity  v = 0

Step 2: Concept of Newton’s 2 nd law of motion

An object will accelerate in the direction of the net force, according to Newton's second law. This acceleration causes the object to slow down and eventually cease moving forward because the force of friction acts in the opposite direction to that of motion.

Step 3: (a) Determining the shortest distance in which car can stop

By applying the Newton’s law to the car

  $$\begin{gathered} ma=-{\mu }_{k}mg \\ a=-{\mu }_{k}g \end{gathered}$$

Substituting the values of ${\mu }_k$ and g

 $$\begin{gathered} a=-0.8\times 9.8m/s^2 \\ =-7.84m/s^2 \end{gathered}$$

Using equation of motion 

 $$\begin{gathered} v^2=u^2+2aS \\ 0=\left(28.7m/s^2\right)-\left(2\times 7.84m/s^2\times S\right) \\ S=\frac{\left(28.7m/s^2\right)}{2\times 7.84m/s^2} \\ =52.53m \end{gathered}$$ 

  

So the shortest distance in which you can stop a car by locking the brakes when the car is traveling at 28.7m/s is 52.53m

Step 4: (b) Determining the initial speed for μ k = 0 . 25 and S = 52.53

If ${\mu }_k=0.25$

By applying the Newton’s law to the car

$$\begin{gathered} ma=-{\mu }_{k}mg \\ a=-{\mu }_{k}g \end{gathered}$$ 

Substituting the values of ${\mu }_k$ and g

$$\begin{gathered} a=0.25\times 9.8m/s^2 \\ =-2.45m/s^2 \end{gathered}$$ 

Using equation of motion

 $$\begin{gathered} v^2=u^2+2aS \\ 0=u^2-\left(2\times 2.45m/s^2\times 52.53m\right) \\ {u}^2=257.397m^2/s^2 \\ u=16.04m/s \end{gathered}$$

So car should be drive at initial velocity  16.04m/s on wet pavement to be able to stop in the same distance as in part (a) when the co-efficient of friction is 0.25