High School

A car is traveling at [tex]$57 \text{ mi/h}$[/tex] before it enters into a skid. The drag factor of the road surface is 1.1, and the braking efficiency is [tex]$100\%$[/tex]. How long might the average skid mark be to the nearest tenth of a foot?

Use the formula: [tex]$s = \sqrt{30Dn}$[/tex]

A. 98.5
B. 95.8
C. 57
D. 57.5

Answer :

To solve this problem, we need to calculate the average skid mark length when a car is traveling at a speed of 57 miles per hour, with a drag factor of 1.1 and a braking efficiency of 100%.

The formula used to find the skid mark length [tex]\( s \)[/tex] is given by:

[tex]\[ s = \sqrt{30 \times D \times n} \][/tex]

where:
- [tex]\( D \)[/tex] is the speed in miles per hour (mi/h).
- [tex]\( n \)[/tex] is the drag factor (a measure of how much the road surface slows down the car).

Given:
- [tex]\( D = 57 \)[/tex] mi/h
- [tex]\( n = 1.1 \)[/tex]
- Braking efficiency is 100%, which means it doesn't change our calculation.

To find the skid mark length [tex]\( s \)[/tex]:

1. Multiply the speed [tex]\( D \)[/tex] by the drag factor [tex]\( n \)[/tex]:
[tex]\[ 57 \times 1.1 = 62.7 \][/tex]

2. Multiply the result by 30:
[tex]\[ 30 \times 62.7 = 1881 \][/tex]

3. Find the square root of the product:
[tex]\[ \sqrt{1881} \approx 43.37 \][/tex]

4. Round the result to the nearest tenth of a foot:
[tex]\[ 43.4 \text{ feet} \][/tex]

Therefore, the average skid mark length, to the nearest tenth of a foot, is [tex]\(\boxed{43.4}\)[/tex] feet.