A driver approaches an intersection at a speed of 50km/h, when the light turns amber. The driver applies brakes to get the maximum stopping force without skidding. The car has a mass of 1500kg and the force of static friction between the tires and the road is 1.1*10^4N. Assuming the driver's reaction time is 0.3s. Find the maximum acceleration of the car, the maximum stopping time, and the minimum stopping distance.

To find the maximum acceleration of the car, we need to determine the maximum force of static friction that the tires can exert. The maximum force of static friction is equal to the product of the coefficient of static friction (μs) and the normal force (N) between the tires and the road.

However, the normal force is equal to the weight of the car, which is the product of the mass (m) and the acceleration due to gravity (g). The equation for the maximum force of static friction is:

Maximum Force of Static Friction (Ff) = μs * N

Normal Force (N) = m * g

Given:
Mass of the car (m) = 1500 kg
Acceleration due to gravity (g) = 9.8 m/s^2
Coefficient of static friction (μs) = force of static friction (1.1 * 10^4 N) / normal force (N)

First, we calculate the normal force:

N = m * g
N = 1500 kg * 9.8 m/s^2
N = 14700 N

Next, we can calculate the maximum force of static friction:

Ff = μs * N
Ff = (1.1 * 10^4 N) / (14700 N)
Ff = 0.748 N

Finally, we can calculate the maximum acceleration:

Using Newton's second law of motion, we can relate force and acceleration:

F = m * a

Rearranging the equation, we get:

a = F / m

a = 0.748 N / 1500 kg
a = 0.0004993 m/s^2

Therefore, the maximum acceleration of the car is approximately 0.0004993 m/s^2.

To find the maximum stopping time, we can use the equation for uniform acceleration:

v = u + at

Where:
u = initial velocity (50 km/h)
v = final velocity (0 m/s)
a = acceleration (calculated above)

First, we need to convert the initial velocity from km/h to m/s:

Initial velocity (u) = 50 km/h * (1000 m/3600 s)
u = 13.89 m/s

Next, we rearrange the equation to solve for time (t):

v = u + at
0 = 13.89 m/s + (0.0004993 m/s^2) * t

Solving for t, we get:

t = -13.89 m/s / (0.0004993 m/s^2)
t = -27,814.64 s

Since the time cannot be negative, we ignore the negative sign:

t = 27,814.64 s

Therefore, the maximum stopping time is approximately 27,814.64 seconds.

To find the minimum stopping distance, we can use the equation of motion:

s = ut + 1/2 * a * t^2

Where:
s = stopping distance (unknown)
u = initial velocity (13.89 m/s)
a = acceleration (0.0004993 m/s^2)
t = stopping time (27,814.64 s)

Plugging in the values, we get:

s = (13.89 m/s * 27,814.64 s) + 1/2 * (0.0004993 m/s^2) * (27,814.64 s)^2

Simplifying the equation, we get:

s = 193,700.46 m + 10,398,102 m

Adding the two terms, we get:

s = 10,591,802.46 m

Therefore, the minimum stopping distance is approximately 10,591,802.46 meters.