Question

A car slams on its brakes while traveling at 80 kph. If the coefficient of friction between the tires and the pavement is 0.6, for how much time does the car skid? What if the pavement is wet, reducing the coefficient to 0.3? What if the road is snow-covered, reducing the coefficient to 0.12?

Answer 1

The car will skid for approximately 5.14 seconds on dry pavement, 10.29 seconds on wet pavement, and 25.73 seconds on snow-covered road.

Step-by-step explanation:

To calculate the time it takes for a car to skid when the brakes are slammed on, we need to use the equation of motion for constant deceleration:

Final velocity (v) = Initial velocity (u) + Acceleration (a) * Time (t)

The initial velocity is the speed of the car before braking, which is 80 kph. The final velocity is 0 kph since the car comes to a stop. The acceleration can be calculated using the coefficient of friction and the formula:

Acceleration (a) = Coefficient of friction (μ) * Acceleration due to gravity (g)

Using these equations, we find that the car will skid for approximately 5.14 seconds on dry pavement. If the pavement is wet, reducing the coefficient of friction to 0.3, the car will skid for approximately 10.29 seconds. If the road is snow-covered, reducing the coefficient of friction to 0.12, the car will skid for approximately 25.73 seconds.

Answer 2

The time it takes for a car to stop skidding can be calculated using the following equation:

t = √(2d / a)

where:

t = time (s)

d = the distance traveled while skidding (m)

a = the acceleration due to friction (m/s^2)

The acceleration due to friction (a) can be calculated as follows:

a = μ * g

where:

μ = the coefficient of friction

g = the acceleration due to gravity (9.8 m/s^2)

Dry pavement (μ = 0.6):

a = 0.6 * 9.8 = 5.88 m/s^2

d = v^2 / (2a) = (80 / 3.6)^2 / (2 * 5.88) = 31.23 m

t = √(2d / a) = √(2 * 31.23 / 5.88) = 2.51 s

Wet pavement (μ = 0.3):

a = 0.3 * 9.8 = 2.94 m/s^2

d = v^2 / (2a) = (80 / 3.6)^2 / (2 * 2.94) = 62.46 m

t = √(2d / a) = √(2 * 62.46 / 2.94) = 4.30 s

Snow-covered pavement (μ = 0.12):

a = 0.12 * 9.8 = 1.17 m/s^2

d = v^2 / (2a) = (80 / 3.6)^2 / (2 * 1.17) = 262.48 m

t = √(2d / a) = √(2 * 262.48 / 1.17) = 9.98 s