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Mark was traveling at 40 mph on a road with a drag factor of 0.85. His brakes were working at B0o/o efficiency. To the nearest tenth of a foot, what would you expect the average length of the skid marks to be if he applied his brakes in order to come to an immediate stop

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The average length of the skid marks to be if he applied his brakes in order to come to an immediate stop is 78.43 ft.

The given parameters:

  • Speed of Mark, v = 40 mph
  • Drag factor, F = 0.85
  • Working efficiency of the brake, n = 80%

The average length of the skid marks to be if he applied his brakes in order to come to an immediate stop is calculated as follows;

[tex]v = \sqrt{30 D fn} \\\\40 = \sqrt{30 \times D \times 0.85 \times 0.8} \\\\40 = \sqrt{20.4 D} \\\\40^2 = 20.4D\\\\1600 = 20.4 D\\\\D = \frac{1600}{20.4} \\\\D = 78.43 \ ft[/tex]

Thus, the average length of the skid marks to be if he applied his brakes in order to come to an immediate stop is 78.43 ft.

Learn more about length of a skid mark here: https://brainly.com/question/16914218

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