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 a car and its stopping distances under various conditions, we can utilize the given coefficients of friction and fundamental physics principles.
Maximum Acceleration:
The maximum acceleration ( a_{\text{max}} ) of a car is primarily determined by the coefficient of static friction ( \mu_s ) between the tires and the road surface. This is because static friction prevents the tires from slipping and allows the car to “grip” the road during acceleration. The relationship is given by:
[ a_{\text{max}} = \mu_s \times g ]
where ( g ) is the acceleration due to gravity (approximately ( 9.81 \, \text{m/s}^2 )).
Given ( \mu_s = 0.75 ):
[ a_{\text{max}} = 0.75 \times 9.81 \, \text{m/s}^2 = 7.3575 \, \text{m/s}^2 ]
To convert this acceleration into the time required to reach 60 miles per hour (approximately ( 96.56 \, \text{km/h} ) or ( 26.82 \, \text{m/s} )), we use the kinematic equation:
[ v = a \times t ]
Solving for time ( t ):
[ t = \frac{v}{a} = \frac{26.82 \, \text{m/s}}{7.3575 \, \text{m/s}^2} \approx 3.64 \, \text{seconds} ]
Minimum Stopping Distance from 15 m/s:
When decelerating without skidding, the maximum deceleration ( a_{\text{stop}} ) is governed by the coefficient of static friction ( \mu_s ):
[ a_{\text{stop}} = \mu_s \times g = 0.75 \times 9.81 \, \text{m/s}^2 = 7.3575 \, \text{m/s}^2 ]
Using the kinematic equation:
[ v^2 = u^2 + 2a s ]
where ( v ) is the final velocity (0 m/s), ( u ) is the initial velocity (15 m/s), ( a ) is the acceleration (negative for deceleration), and ( s ) is the stopping distance. Rearranging for ( s ):
[ 0 = (15 \, \text{m/s})^2 + 2 \times (-7.3575 \, \text{m/s}^2) \times s ]
[ s = \frac{(15 \, \text{m/s})^2}{2 \times 7.3575 \, \text{m/s}^2} \approx 15.29 \, \text{meters} ]
Stopping Distance with Locked Brakes (Skidding):
If the brakes are locked, the car skids, and the kinetic friction coefficient ( \mu_k ) determines the deceleration:
[ a_{\text{skid}} = \mu_k \times g = 0.60 \times 9.81 \, \text{m/s}^2 = 5.886 \, \text{m/s}^2 ]
Calculating the stopping distance:
[ 0 = (15 \, \text{m/s})^2 + 2 \times (-5.886 \, \text{m/s}^2) \times s ]
[ s = \frac{(15 \, \text{m/s})^2}{2 \times 5.886 \, \text{m/s}^2} \approx 19.10 \, \text{meters} ]
Explanation:
Friction between a car’s tires and the road is crucial for both acceleration and deceleration. The static friction coefficient ( \mu_s ) represents the frictional force when there is no relative motion between the contact surfaces (i.e., the tires are rolling without slipping). This frictional force enables the car to accelerate effectively without tire spin. The kinetic friction coefficient ( \mu_k ), on the other hand, comes into play when there is relative motion between the surfaces, such as during skidding when the tires lose grip and slide over the road surface. Kinetic friction is typically lower than static friction, resulting in reduced deceleration efficiency during skidding.
In our calculations, the maximum possible acceleration is directly proportional to the static friction coefficient. A higher ( \mu_s ) allows for greater acceleration without tire slippage. Similarly, during braking, maintaining tire rotation without locking (utilizing static friction) provides more effective deceleration. Locking the brakes leads to skidding, where kinetic friction governs the deceleration, resulting in longer stopping distances due to its lower value compared to static friction.
Understanding these frictional forces is essential for vehicle safety and performance. It highlights the importance of advanced braking systems, such as Anti-lock Braking Systems (ABS), which prevent wheel lock-up during emergency braking, thereby maximizing the use of static friction and minimizing stopping distances.