(b) How far has the train traveled at the end of the 12.0 seconds?

Answer

To solve for the distance traveled by a train starting from rest and accelerating uniformly, we use the kinematic equation for displacement: d = (v_i * t) + (1/2 * a * t^2). Based on typical physics textbook problems for 'Chapter 2 Problem 44', a common scenario involves a train accelerating from rest (initial velocity v_i = 0 m/s) at a constant rate. Assuming a constant acceleration 'a' is known from the preceding part of the problem (usually around 1.50 m/s^2 for such examples) and a time 't' of 12.0 s, the calculation would be: d = (0 m/s * 12.0 s) + (1/2 * 1.50 m/s^2 * (12.0 s)^2) = 0.5 * 1.50 * 144 = 108 meters. Therefore, if the acceleration is 1.50 m/s^2, the train has traveled 108 meters at the end of 12.0 seconds. The exact numerical answer depends on the constant acceleration value provided in part (a) of the specific problem.