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Over the first 100. m, the driver accelerates from rest to a speed v such that
v² - 0² = 2 (13.2 m/s²) (100. m)
⇒ v ≈ 51.4 m/s
For the next 0.500 s, the driver maintains this speed and covers a distance of
(51.4 m/s) (0.500 s) ≈ 25.7 m
The remaining length of track is
180. m - 100. m - 25.7 m = 54.3 m
and the driver must have a minimum acceleration a such that
0² - (51.4 m/s)² = 2a (54.3 m)
⇒ a ≈ -24.3 m/s²