Question

question 16 (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: Identify the physics concept

This problem involves the work - energy principle. The work done by friction to stop the car is equal to the initial kinetic energy of the car. The force of friction is given by \(F_f=\mu_k F_N\), and on a horizontal surface, \(F_N = mg\), so \(F_f=\mu_k mg\). The work done by friction \(W = - F_f d\) (negative because the force is opposite to the displacement), and the initial kinetic energy is \(KE=\frac{1}{2}mv^2\). Setting \(W = \Delta KE\), and since the final velocity \(v_f = 0\), we have \(-\mu_k mgd=0 - \frac{1}{2}mv^2\). The mass \(m\) cancels out.

Step2: Derive the formula for initial velocity

From \(-\mu_k mgd=-\frac{1}{2}mv^2\), we can cancel the negative signs and the mass \(m\) from both sides. We get the formula for the initial velocity \(v=\sqrt{2\mu_k gd}\), where \(\mu_k = 0.500\), \(g = 9.8\ m/s^2\), and \(d = 90.0\ m\).

Step3: Substitute the values into the formula

Substitute \(\mu_k = 0.500\), \(g = 9.8\ m/s^2\), and \(d=90.0\ m\) into the formula \(v=\sqrt{2\times0.500\times9.8\times90.0}\). First, calculate the product inside the square root: \(2\times0.500\times9.8\times90.0=9.8\times90.0 = 882\). Then, take the square root of 882: \(v=\sqrt{882}\approx29.7\ m/s\).

Answer:

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