A policeman investigating an accident measures the skid marks left by a car on the horizontal road. He determines that the distance between the point that the driver slammed on the brakes
(thereby locking the wheels) and the point where the car came to a stop was 28.0 m. From a
Reference manual he determines that the coefficient of kinetic friction between the tires and the
Road under the prevailing conditions was 0.300. How fast was the car going when the driver
Applied the brakes?
A) 32.9 m/s
B) 12.8 m/s
C) 10.7 m/s
D) 45.7 m/s
E) 21.4 m/s

Question

Quizplus · Accepted Answer

The speed of the car when the brakes were applied can be calculated using the formula derived from the work-energy principle, $v = \sqrt{2ad}$, where $v$ is the velocity, $a$ is the acceleration (deceleration in this case), and $d$ is the distance. The deceleration $a$ can be found using the formula $a = \mu g$, where $\mu$ is the coefficient of kinetic friction (0.300) and $g$ is the acceleration due to gravity (approximately 9.8 m/s$^2$). Substituting the given values, $a = 0.300 \times 9.8 = 2.94$ m/s$^2$, and then calculating the velocity gives $v = \sqrt{2 \times 2.94 \times 28.0} \approx 12.8$ m/s.

A Policeman Investigating an Accident Measures the Skid Marks Left