Final answer:
The acceleration of the train as it slows down over a 9.00-s interval is -1.00 m/s², which indicates a deceleration since the final velocity is less than the initial velocity.
Step-by-step explanation:
The acceleration of the train during the 9.00-s interval can be found by using the equation for acceleration, which is a = (v_f - v_i) / t, where v_f is the final velocity, v_i is the initial velocity, and t is the time interval. In this case, the train slows down from an initial velocity of 12.0 m/s to a final velocity of 3.00 m/s over a time of 9.00 s.
To calculate the acceleration, we subtract the final velocity from the initial velocity and then divide by the time interval:
a = (3.00 m/s - 12.0 m/s) / 9.00 s
a = -9.00 m/s / 9.00 s
a = -1.00 m/s²
The negative sign indicates that the train is decelerating, or slowing down, which is consistent with the direction of the acceleration being opposite to the direction of motion. Therefore, the acceleration of the train is -1.00 m/s².