Question

question 3 (1 point)
during the investigation of a traffic accident, police find skid marks 90.0 m long. they determine the coefficient of friction between the cars tires and the roadway to be 0.500 for the prevailing conditions. estimate the speed of the car when the brakes were applied.
9.49 m/s
21.0 m/s
29.7 m/s
42.0 m/s

Explanation:

Step1: Determine the deceleration

The frictional force provides the deceleration. The frictional force \( F_f=\mu N \), and for a car on a horizontal road, \( N = mg \) (normal force equals weight). By Newton's second law \( F = ma \), so the deceleration \( a=\frac{F_f}{m}=\mu g \). Here, \( \mu = 0.500 \) and \( g = 9.8\ m/s^2 \), so \( a=\mu g=0.500\times9.8\ m/s^2 = 4.9\ m/s^2 \).

Step2: Use kinematic equation

We use the kinematic equation \( v_f^2=v_i^2 + 2ax \), where \( v_f = 0\ m/s \) (final velocity when the car stops), \( a=- 4.9\ m/s^2 \) (negative because it's deceleration), and \( x = 90.0\ m \). Rearranging for \( v_i \), we get \( v_i=\sqrt{- 2ax} \). Substituting the values: \( v_i=\sqrt{-2\times(- 4.9\ m/s^2)\times90.0\ m}=\sqrt{882\ m^2/s^2}\approx29.7\ m/s \).

Answer:

29.7 m/s (corresponding to the option "29.7 m/s")