Final answer:
The object decelerates from an initial speed of 54.0 m/s to a stop with a constant acceleration of -6.80 m/s². By using the kinematic equation, we can calculate the distance it travels during this period to be 216.0 meters.
Step-by-step explanation:
The student's question involves calculating the distance an object travels while decelerating to a stop from an initial velocity, with a given acceleration. To solve this, we can use a kinematic equation that relates initial velocity, acceleration, and distance. Specifically, the kinematic equation we use is:
v^2 = u^2 + 2as (where v = final velocity, u = initial velocity, a = acceleration, and s = distance traveled).
Given that the final velocity v is 0 m/s (since the object comes to a complete stop), the initial velocity u is 54.0 m/s, and the acceleration a is -6.80 m/s² (negative because it is decelerating), we can rearrange the formula to solve for s (the distance traveled):
0 = (54.0 m/s)² + 2(-6.80 m/s²)s
Solving for s gives us:
s = - (54.0 m/s)² / (2 * -6.80 m/s²)
s = 216.0 m
Therefore, the object travels a distance of 216.0 meters while decelerating to a stop.