A car travleing at 30.6 mi/h on a horizontal highway. The acceleration of gravity is 9.8 m/s^2. If the coefficient of friction between the road and tires on a rainy day is .083, what is the minimum distance the car will stop?
Work done against friction = Initial K.E.
Ff*X = (M/2)Vo^2 = M*g*0.083 *X
The M's cancel.
Solve for X
To determine the minimum distance the car will stop, we need to analyze the forces acting on the car and apply the laws of motion.
The first step is to convert the speed of the car from miles per hour (mi/h) to meters per second (m/s). Since 1 mile is approximately equal to 1.60934 kilometers and 1 kilometer is equal to 1000 meters, we can use the conversion factor:
30.6 mi/h × (1.60934 km/mi) × (1000 m/km) ÷ (3600 s/h) ≈ 13.67 m/s
Now, let's identify the forces acting on the car:
1. The force of gravity, which is acting vertically downward.
2. The normal force, which is the force exerted by the road on the car and is equal in magnitude and opposite in direction to the force of gravity.
3. The frictional force, which opposes the motion of the car and is responsible for bringing it to a stop.
The frictional force can be calculated using the equation:
frictional force = coefficient of friction × normal force
The normal force can be calculated as:
normal force = mass × acceleration due to gravity
Now, using Newton's second law of motion:
frictional force = mass × acceleration
In this case, we want to determine the minimum distance the car will stop, which means it will come to rest. This implies that the net force acting on the car is zero. Hence, the frictional force is equal in magnitude to the force of gravity:
frictional force = force of gravity
Substituting the equations above, we have:
coefficient of friction × normal force = mass × acceleration
coefficient of friction × (mass × acceleration due to gravity) = mass × acceleration
Simplifying and canceling out the mass term from both sides:
coefficient of friction × acceleration due to gravity = acceleration
Now we can substitute the given values:
0.083 × 9.8 m/s^2 = acceleration
acceleration ≈ 0.814 m/s^2
Now, to calculate the minimum distance the car will stop, we can use the following kinematic equation:
vf^2 = vi^2 + 2as
Where:
vf = final velocity (0 m/s since the car comes to a stop)
vi = initial velocity (13.67 m/s)
a = acceleration (-0.814 m/s^2, negative since it's opposite to the initial velocity)
s = distance
Substituting the values:
0^2 = (13.67 m/s)^2 + 2(-0.814 m/s^2)s
Simplifying the equation:
0 = 187.28 m^2/s^2 - 1.628 s^2
Rearranging the equation:
1.628 s^2 = 187.28 m^2/s^2
Now, solving for s:
s^2 = 187.28 m^2/s^2 / 1.628
s^2 ≈ 115.03 m^2/s^2
Taking the square root of both sides:
s ≈ √115.03 m
s ≈ 10.73 m
Therefore, the minimum distance the car will stop is approximately 10.73 meters.