Final answer:
The maximum speed of the cyclist after accelerating at 2 m/s² for 10 seconds is 20 m/s, and the total distance covered, including acceleration, constant speed, and deceleration phases, is 1366.7 meters. None of the options provided by the student exactly match this answer.
Step-by-step explanation:
The student is solving a physics problem involving the motion of a cyclist in three stages: acceleration, constant velocity, and deceleration. To determine the maximum speed and total distance covered, we need to apply the equations of motion for each stage.
Acceleration stage:
- Maximum speed after accelerating for 10 seconds: v = u + at, where u = 0 m/s (initial velocity), a = 2 m/s², and t = 10 s. So the maximum speed v = 0 + (2 m/s² × 10 s) = 20 m/s.
- Distance covered while accelerating: d1 = ut + (1/2)at² = 0 + (1/2)(2 m/s²)(10 s)² = 100 m.
Constant speed stage:
- Distance covered at constant speed for one minute (60 seconds): d2 = vt = 20 m/s × 60 s = 1200 m.
Deceleration stage:
- Using v = u + at to find the time to stop, with v = 0 m/s, u = 20 m/s and a = -3 m/s²: 0 = 20 m/s - (3 m/s²)t. Hence, t = 20 m/s / 3 m/s² = 6.67 s.
- Distance covered while decelerating: d3 = ut + (1/2)at² = (20 m/s × 6.67 s) + (1/2)(-3 m/s²)(6.67 s)² = 133.4 m - 66.7 m = 66.7 m.
Total distance covered: d_total = d1 + d2 + d3 = 100 m + 1200 m + 66.7 m = 1366.7 m.
Option (a) fits the calculated maximum speed after acceleration, but the calculated total distance is not listed in the options provided by the student. Therefore, none of the options perfectly match the correct answer.
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