An ice skater traveling in a straight line along flat ice with initial speed $v$ slows to a stop after traveling $12.3 \mathrm{~m}$. If the coefficient of kinetic friction between the ice skater and the ice is 0.070 , what was the skater's initial speed $v$ ? You may ignore air resistance, and the acceleration due to gravity is $a_{y}=-g$. $4.11 \mathrm{~m} / \mathrm{s}$ $3.55 \mathrm{~m} / \mathrm{s}$ $2.90 \mathrm{~m} / \mathrm{s}$ $3.17 \mathrm{~m} / \mathrm{s}$

Step 1 ：We are given that an ice skater traveling in a straight line along flat ice with initial speed \(v\) slows to a stop after traveling \(12.3 \mathrm{~m}\). The coefficient of kinetic friction between the ice skater and the ice is \(0.070\). We are asked to find the skater's initial speed \(v\).

Step 2 ：We can use the equation of motion to solve this problem. The equation of motion is given by \(v^2 = u^2 + 2as\), where \(v\) is the final velocity, \(u\) is the initial velocity, \(a\) is the acceleration and \(s\) is the distance.

Step 3 ：In this case, the final velocity \(v\) is 0 (since the skater comes to a stop), the distance \(s\) is 12.3 m, and the acceleration \(a\) is the frictional force divided by the mass of the skater.

Step 4 ：The frictional force is given by \(f = \mu m g\), where \(\mu\) is the coefficient of friction, \(m\) is the mass of the skater, and \(g\) is the acceleration due to gravity. Since the frictional force acts in the opposite direction to the motion of the skater, the acceleration \(a\) is negative.

Step 5 ：Therefore, we can write the equation of motion as \(0 = u^2 - 2 \mu g s\). Solving for \(u\), we get \(u = \sqrt{2 \mu g s}\).

Step 6 ：Substituting the given values into the equation, we get \(u = \sqrt{2 \times 0.07 \times 9.81 \times 12.3}\).

Step 7 ：Calculating the above expression, we get \(u = 4.11 \mathrm{~m/s}\).

Step 8 ：\(\boxed{4.11 \mathrm{~m/s}}\) is the initial speed of the skater.