Answer:
Policeman is correct.
Step-by-step explanation:
Consider the provided information.
The tires left three skid marks of 69 ft, 70 ft, and 74 ft.
The road had a drag factor of 0.95. Your brakes were operating at 98% efficiency.
Thus, drag factor (f) = 0.95 and brakes efficiency (n) = 0.98
Compute the skid distance by finding the mean of length of 3 skids.
[tex]D=\frac{69+70+74}{3}\\D=\frac{213}{3}\\D=71[/tex]
Therefore, the skid distance is 71 ft.
Now find the speed of car using skid speed formula:
[tex]S=\sqrt{30Dfn}[/tex]
Where, S is the speed of car, D is the skid distance, f is the drag factor and n is the beak efficiency.
Substitute the respective values in above formula.
[tex]S=\sqrt{30\times 71\times0.95 \times0.98}[/tex]
[tex]S=\sqrt{1983.03}[/tex]
[tex]S\approx 44.5312[/tex]
Your speed was 44.5312 mi/h which is greater than 40 mi/hr. So policeman is correct.
