A car skids to a stop, leaving four skid marks with an average length (D) of 60 feet. The road is asphalt. Skid tests reveal a drag factor(f) of 0.75. Since all four wheels were braking, the braking efficiency (n) is 100% or 1.00. To the nearest mile per hour, what was the car’s speed at the time of the accident
To determine the car's speed at the time of the accident, we can use the following formula:
v = sqrt((2 * f * n * g * D) / ((1 - n) * C))
Where:
v = Speed of the car
f = Drag factor
n = Braking efficiency
g = Acceleration due to gravity (32.174 ft/s^2)
D = Average length of skid marks
C = Coefficient of friction
In this case, the road is asphalt, and the average coefficient of friction for asphalt is around 0.7.
Let's calculate the car's speed:
v = sqrt((2 * 0.75 * 1.00 * 32.174 * 60) / ((1 - 1.00) * 0.7))
v = sqrt((2 * 0.75 * 32.174 * 60) / (0.7))
Using a calculator, we find that v ≈ 65.76 ft/s.
To convert this speed to miles per hour (mph), we multiply by the conversion factor:
v_mph = v * 0.681818
v_mph ≈ 44.93 mph
Therefore, to the nearest mile per hour, the car's speed at the time of the accident was approximately 45 mph.
To determine the car's speed at the time of the accident, we can use the following formula:
Speed (v) = (D * D * f * n) / 30
Given:
- Average length of skid marks (D) = 60 feet
- Drag factor (f) = 0.75
- Braking efficiency (n) = 1.00
We can substitute these values into the formula:
v = (60 * 60 * 0.75 * 1.00) / 30
Calculating this expression:
v = 2700 / 30
v = 90 feet per second
Since we want to determine the speed in miles per hour, we need to convert feet per second to miles per hour. There are 5280 feet in a mile and 3600 seconds in an hour.
v = (90 * 3600) / (5280 * 1)
v = 54,000 / 5280
v ≈ 10.23 miles per hour
Rounding to the nearest whole number, the car's speed at the time of the accident was approximately 10 miles per hour.
(1/2) m v^2 = force * distance = .75 weight * 60 feet
m = weight/g = weight in pounds/32 approx
(1/64) v^2 = .75 * 60
v is in ft/s, convert to mph