A 1400 kg car moves along a horizontal road at speed v0 = 21.4 m/s. The road is wet, so the static friction coefficient between the tires and the road is only μs = 0.127 and the kinetic friction coefficient is even lower, μk = 0.0889.

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 determine the maximum static friction force that can be applied to the car to bring it to a stop.

The maximum static friction force (fs) can be calculated using the equation:

fs = μs * m * g

where μs is the static friction coefficient (0.127), m is the mass of the car (1400 kg), and g is the acceleration due to gravity (9.8 m/s^2).

Plugging in the values:
fs = 0.127 * 1400 kg * 9.8 m/s^2 = 1721.696 N

Now, we can calculate the acceleration of the car using the maximum static friction force:

fs = m * a
1721.696 N = 1400 kg * a
a = 1.229056 m/s^2

Next, we can use the kinematic equation to find the stopping distance (s) of the car:

v^2 = u^2 + 2as

where v is the final velocity (0 m/s), u is the initial velocity (21.4 m/s), a is the acceleration (-1.229056 m/s^2), and s is the stopping distance.

Plugging in the values:
0 = (21.4 m/s)^2 + 2 * (-1.229056 m/s^2) * s

Rearranging the equation:
s = (21.4 m/s)^2 / (2 * 1.229056 m/s^2)

Calculating:
s = 232.6865 m

Therefore, the shortest possible stopping distance for the car is approximately 232.6865 meters.

To find the shortest possible stopping distance for the car, we need to determine the maximum deceleration that can be achieved based on the coefficient of kinetic friction and the car's mass.

The maximum deceleration can be calculated using the formula:

a = μk * g

where a is the maximum deceleration, μk is the coefficient of kinetic friction (0.0889), and g is the acceleration due to gravity (9.8 m/s^2).

Substituting the given values, we have:

a = 0.0889 * 9.8
a ≈ 0.86922 m/s^2

Next, we can use the kinematic equation to calculate the stopping distance:

v^2 = u^2 + 2as

where v is the final velocity (0 m/s since the car is stopping), u is the initial velocity (21.4 m/s), a is the acceleration, and s is the stopping distance.

Rearranging the equation to solve for s, we have:

s = (v^2 - u^2) / (2a)

Substituting the given values, we get:

s = (0 - 21.4^2) / (2 * 0.86922)
s ≈ -21.4^2 / (2 * 0.86922)
s ≈ -459.56 / 1.73844
s ≈ -264.68 m

However, it is important to note that we are dealing with the physical quantity of distance, which cannot be negative. Therefore, we disregard the negative sign and take the absolute value of the calculated distance.

s ≈ |-264.68| ≈ 264.68 m

Hence, the shortest possible stopping distance for the car under these conditions is approximately 264.68 meters.

Wc = M*g = 1400 * 9.8 = 13,720 N. = Normal force, Fn.

Fp = mg*sin 0 = 0. = Force parallel to
the road.

Fk = u*Fn = 0.0889 * 13,720 = 1220 N.

a = (Fp-Fk)/M = (0-1220)/1400 = -0.871 m/s^2.

Vf^2 = Vo^2 + 2a*d.
Vf = 0.
a = -0.871 m/s^2.
Vo = 21.4 m/s.
d = ?.