A 940 kg car moves along a horizontal road at speed v0 = 16.8 m/s. The road is wet, so the static friction coefficient between the tires and the road is only μs = 0.196 and the kinetic friction coefficient is even lower, μk = 0.1372.
The acceleration of gravity is 9.8 m/s2 .
What is the shortest possible stopping dis- tance for the car under such conditions? Use g = 9.8 m/s2 and neglect the reaction time of the driver.
Answer in units of m
To find the shortest possible stopping distance for the car, we need to first calculate the maximum static friction force that can act on the car to bring it to a stop.
The maximum static friction force (F_s) can be calculated using the formula:
F_s = μ_s * m * g
where:
μ_s is the static friction coefficient (0.196),
m is the mass of the car (940 kg),
and g is the acceleration due to gravity (9.8 m/s^2).
Plugging in the given values into the formula, we get:
F_s = 0.196 * 940 kg * 9.8 m/s^2
F_s = 1764.72 N (rounded to 2 decimal places)
Next, we can calculate the maximum deceleration (a_s) the car can experience due to the maximum static friction force using Newton's second law of motion:
F_s = m * a_s
Solving for a_s, we have:
a_s = F_s / m
a_s = 1764.72 N / 940 kg
a_s = 1.87755 m/s^2 (rounded to 5 decimal places)
Now, we can calculate the stopping time (t_s) for the car using the formula:
t_s = v_0 / a_s
where v_0 is the initial velocity of the car (16.8 m/s).
Plugging in the values, we get:
t_s = 16.8 m/s / 1.87755 m/s^2
t_s = 8.94378 s (rounded to 5 decimal places)
Finally, we can calculate the shortest possible stopping distance (d_s) for the car using the formula:
d_s = v_0 * t_s - (1/2) * a_s * t_s^2
Plugging in the values, we get:
d_s = 16.8 m/s * 8.94378 s - (1/2) * 1.87755 m/s^2 * (8.94378 s)^2
d_s = 150.04224 m - 71.015708 m
d_s = 79.026532 m (rounded to 6 decimal places)
Therefore, the shortest possible stopping distance for the car is approximately 79.03 m.
To find the shortest possible stopping distance for the car under the given conditions, we need to determine the stopping force and then use it to calculate the distance traveled during deceleration.
The frictional force between the tires and the road can be represented by the equation:
F_friction = μ * m * g,
where F_friction is the frictional force, μ is the coefficient of friction, m is the mass of the car, and g is the acceleration due to gravity.
In this case, since the car is already moving, we need to consider the kinetic friction coefficient μk, so the frictional force during deceleration will be:
F_friction = μk * m * g.
The net force acting on the car during deceleration is the frictional force F_friction, which opposes the direction of motion. So, the net force is given by:
F_net = -F_friction = -μk * m * g.
Using Newton's second law, F_net = m * a, where a is the acceleration, we can solve for the acceleration:
a = -μk * g.
The car starts at speed v0 and comes to rest when its final velocity vf is zero. We can relate the initial velocity, final velocity, acceleration, and distance using the equation:
vf^2 = v0^2 + 2 * a * d,
where vf is the final velocity, v0 is the initial velocity, a is the acceleration, and d is the stopping distance.
Since vf = 0 and a = -μk * g, the equation simplifies to:
0 = v0^2 - 2 * μk * g * d.
Solving for d, we get:
d = v0^2 / (2 * μk * g).
Substituting the given values, we have:
v0 = 16.8 m/s,
μk = 0.1372,
g = 9.8 m/s^2.
Plugging these values into the equation, we can calculate the shortest possible stopping distance for the car:
d = (16.8 m/s)^2 / (2 * 0.1372 * 9.8 m/s^2).
Calculating this equation, we find:
d ≈ 92.556 m.
Therefore, the shortest possible stopping distance for the car under the given conditions is approximately 92.556 meters.
uk = a/g = 0.1372
a = 0.1372 * g = 0.1372 * (-9.8) =-1.34
m/s^2.
V^2 = Vo^2 + 2a*d = 0 @ max. ht.
d = -(Vo^2)/2a = -(16.8^2)/-1.34 = 210 m