The coefficient of kinetic friction between a particular car’s tires and the ground is 0.60. The coefficient of static friction is 0.75.
Given the above information and the assumption that friction is the limiting factor, not the car’s engine, determine the maximum acceleration of the car. Use this to determine the time to accelerate from zero to 60 miles per hour. (1 mile = 1.61 km)
Determine the minimum stopping distance of the car if it is moving at a speed of 15 m/s. Repeat assuming the driver locks the brakes and skids to a stop.
The correct answer and explanation is :
To determine the maximum acceleration of the car, we utilize the coefficient of static friction (μₛ), as it represents the frictional force preventing the tires from slipping during acceleration. The maximum frictional force (fₛ) can be expressed as:
[ fₛ = μₛ \times N ]
Here, N denotes the normal force, which equals the car’s weight (mass m multiplied by gravitational acceleration g):
[ N = m \times g ]
Substituting this into the frictional force equation:
[ fₛ = μₛ \times m \times g ]
According to Newton’s second law, force equals mass times acceleration (f = m × a). Setting the frictional force equal to the product of mass and acceleration:
[ m \times a = μₛ \times m \times g ]
By canceling the mass (m) from both sides, we find the maximum acceleration (aₘₐₓ):
[ aₘₐₓ = μₛ \times g ]
Given μₛ = 0.75 and g ≈ 9.81 m/s²:
[ aₘₐₓ = 0.75 \times 9.81 \, \text{m/s}^2 \approx 7.36 \, \text{m/s}^2 ]
Next, we calculate the time required to accelerate from 0 to 60 miles per hour (mph). First, convert 60 mph to meters per second (m/s):
[ 60 \, \text{mph} = 60 \times 1.61 \, \text{km/h} = 96.6 \, \text{km/h} ]
Since 1 km/h ≈ 0.27778 m/s:
[ 96.6 \, \text{km/h} \times 0.27778 \, \text{m/s per km/h} \approx 26.83 \, \text{m/s} ]
Using the formula for acceleration (a = Δv / Δt), we solve for time (Δt):
[ Δt = Δv / aₘₐₓ = 26.83 \, \text{m/s} / 7.36 \, \text{m/s}^2 \approx 3.64 \, \text{seconds} ]
For the minimum stopping distance from an initial speed (v₀) of 15 m/s, we apply the work-energy principle. The work done by friction (which brings the car to a stop) equals the initial kinetic energy:
[ fₖ \times d = \frac{1}{2} m v₀^2 ]
Here, fₖ is the kinetic frictional force:
[ fₖ = μₖ \times N = μₖ \times m \times g ]
Solving for the stopping distance (d):
[ μₖ \times m \times g \times d = \frac{1}{2} m v₀^2 ]
Canceling mass (m) from both sides:
[ d = \frac{v₀^2}{2 \times μₖ \times g} ]
Given μₖ = 0.60:
[ d = \frac{(15 \, \text{m/s})^2}{2 \times 0.60 \times 9.81 \, \text{m/s}^2} \approx 19.1 \, \text{meters} ]
If the driver locks the brakes, causing the car to skid, the coefficient of kinetic friction (μₖ) applies. Repeating the calculation with μₖ = 0.60:
[ d = \frac{(15 \, \text{m/s})^2}{2 \times 0.60 \times 9.81 \, \text{m/s}^2} \approx 19.1 \, \text{meters} ]
In both scenarios, the stopping distance is approximately 19.1 meters. This demonstrates that, under the given conditions, locking the brakes does not significantly alter the stopping distance compared to controlled braking at the threshold of static friction.
Understanding these calculations is crucial for vehicle safety and performance. The coefficient of friction between tires and the road surface directly influences acceleration and braking capabilities. A higher coefficient allows for greater acceleration and shorter stopping distances, enhancing overall vehicle control.
It’s important to note that these calculations assume ideal conditions, such as uniform friction and no additional forces (e.g., air resistance). In real-world scenarios, factors like road conditions, tire wear, and vehicle dynamics can affect these outcomes.
In summary, the maximum acceleration of the car is approximately 7.36 m/s², allowing it to accelerate from 0 to 60 mph in about 3.64 seconds. The minimum stopping distance from 15 m/s is around 19.1 meters, regardless of whether the brakes are locked or not. These insights underscore the critical role of friction in vehicle dynamics and the importance of maintaining optimal tire and road conditions for safety.
For a visual explanation of similar concepts, you might find the following video helpful:
videoWhat is the maximum acceleration a car can undergo if the coefficient of static friction between the tires and the ground is 0.90?turn0search0