Final answer:
The runner travels 20 meters in the next 5.00 seconds after beginning to decelerate from a velocity of 9.00 m/s. Part (a) of the problem yields the distance traveled, while part (b)'s suggested final velocity indicates an unrealistic scenario of moving backwards.
Step-by-step explanation:
To determine how far the runner travels after decelerating from a velocity of 9.00 m/s at a rate of 2.00 m/s² over 5.00 seconds, we can use the formula for distance covered under constant acceleration, which is d = ut + (1/2)at², where d stands for distance, u is the initial velocity, a is the acceleration, and t is the time.
Substituting the given values, we get:
d = (9 m/s)(5 s) + (1/2)(-2 m/s²)(5 s)² = 45 m - 25 m = 20 m.
So, the runner travels 20 meters in the next 5.00 seconds after beginning to decelerate. Option (a) would be the correct answer. However, part (b)'s result, a final velocity of -1.00 m/s, is not physically realistic for the scenario, as it suggests that the runner would be moving backwards after coming to a stop. The runner would actually come to a stop after 4.50 seconds, as the deceleration would have negated the initial velocity by then ((9 m/s) / (2 m/s²) = 4.5 s).