Problem 36 - College Physics

Question

\(\bullet$$\bullet\) Stopping distance of a car. (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 an automobile by lock- ing the brakes when traveling at 29.1 \(\mathrm{m} / \mathrm{s}\) (about 65 \(\mathrm{mi} / \mathrm{h} )\) ? (b) On wet pavement, the coefficient of kinetic friction may be only \(0.25 .\) How fast should you drive on wet pavement in ordel 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) 54 meters; (b) Drive at 16.3 m/s on wet pavement.

Step 1: Understand Given Information

We need to find the stopping distance for an automobile under two different friction conditions. In part (a), the coefficient of kinetic friction on dry pavement is given as \(\mu = 0.80\) and the initial speed is \(v_0 = 29.1\, \text{m/s}\). In part (b), the coefficient of kinetic friction on wet pavement is \(\mu = 0.25\). We have to find the speed such that the stopping distance is the same as in part (a).

Step 2: Determine Deceleration Due to Friction

The deceleration due to friction \(a\) can be calculated using the formula \(a = \mu g\), where \(g = 9.8 \, \text{m/s}^2\) is the acceleration due to gravity. For part (a):\[ a = 0.80 \times 9.8 = 7.84 \, \text{m/s}^2. \] In part (b), we only need this value to keep calculations similar.

Step 3: Use Kinematic Equation to Calculate Stopping Distance

The stopping distance \(d\) can be determined using the kinematic equation \(v^2 = v_0^2 + 2ad\). Since the final velocity \(v = 0\) when the car stops: \[ 0 = (29.1)^2 + 2(-7.84)d. \] Solving this for \(d\), we get: \[ d = \frac{(29.1)^2}{2(7.84)} = 54 \text{ m}. \]

Step 4: Calculate Initial Speed on Wet Pavement

For part (b), we use the same kinematic equation but adjust it for the new \(\mu = 0.25\). Since the stopping distance \(d\) is 54 m, we have \(a = 0.25 \times 9.8 = 2.45 \, \text{m/s}^2\): \[ 0 = v_{wet}^2 - 2(2.45)(54). \] Solving for \(v_{wet}\), we find: \[ v_{wet} = \sqrt{2 \times 2.45 \times 54} \approx 16.3 \, \text{m/s}. \]

Step 5: Finalize the Results

In summary, the shortest stopping distance on dry pavement with a friction coefficient of 0.80 is 54 meters. To achieve this stopping distance on wet pavement (friction coefficient 0.25), the driving speed should be reduced to approximately 16.3 m/s.

Key Concepts

Kinetic Friction

Kinetic friction is the force that resists the movement of two objects sliding against each other. In the context of stopping a car, it is a crucial factor because it determines how effectively the car can come to a halt.
Unlike static friction, which acts on objects that are not sliding, kinetic friction acts when there is relative motion between the surfaces. The kinetic friction force can be calculated using the formula:

• \( f_k = \mu_k N \)

where \( f_k \) is the kinetic friction force, \( \mu_k \) is the coefficient of kinetic friction, and \( N \) is the normal force, typically the weight of the car due to gravity.
When the brakes are locked, the car skids, and kinetic friction becomes the dominant force opposing the car's motion. This friction force is directly proportional to the coefficient of kinetic friction, meaning the higher the value, the greater the resistance, shortening the stopping distance.

Deceleration

Deceleration is essentially negative acceleration, which means a decrease in speed. When a car brakes, it experiences deceleration due to the force of friction acting against its motion.
This force leads to a rapid drop in speed, eventually bringing the vehicle to a stop. The rate of deceleration due to friction can be calculated using the formula:

• \( a = \mu g \)

In this formula, \( a \) is the deceleration, \( \mu \) is the coefficient of friction, and \( g \) is the acceleration due to gravity (about 9.8 m/s² on Earth).
A higher coefficient of friction leads to a greater deceleration, as seen on dry pavement with a friction coefficient of 0.80, resulting in an effective deceleration that allows for shorter stopping distances, while a lower coefficient seen on wet roads leads to decreased deceleration.

Kinematic Equations

Kinematic equations are fundamental in physics for describing the motion of objects. They connect concepts like displacement, initial velocity, final velocity, acceleration, and time.
In the case of stopping a car, the most relevant kinematic equation is:

• \( v^2 = v_0^2 + 2ad \)

where \( v \) is the final velocity (zero when the car stops), \( v_0 \) is the initial velocity, \( a \) is acceleration (or deceleration in this context), and \( d \) is the stopping distance.
This equation is used to calculate how far the car will travel until it completely stops given a certain deceleration. By substituting known values, such as initial speed and deceleration, one can solve for the stopping distance or the necessary initial speed to achieve a particular stopping distance under different conditions.

Coefficient of Friction

The coefficient of friction is a dimensionless number that represents the frictional properties between two surfaces.
There are two main types: static and kinetic, but here, kinetic friction is specially considered, as it applies to objects in motion.
A higher coefficient of kinetic friction indicates that more force is required to keep the motion going, which is helpful in stopping a car quickly on dry pavement. The coefficient is affected by:

• Type of materials in contact (e.g., rubber tires on asphalt)
• Surface conditions (e.g., dry vs. wet pavement)
• Presence of lubricants or contaminants

In this exercise, dry pavement has a coefficient of 0.80, indicating stronger frictional force compared to wet pavement with a coefficient of 0.25. As a result, more force is needed to stop a car on dry pavement, which allows for shorter stopping distances.