a) Accelerating from rest for 20 s, the cyclist would attain a speed of
(0.6 m/s²) (20 s) = 12 m/s
b) In this time, the cylclist would cover a distance of
1/2 (0.6 m/s²) (20 s)² = 120 m
Alternatively, since we know he gets to final speed of 12 m/s and that acceleration is constant, we have
(12 m/s)² - 0² = 2 (0.6 m/s²) x
===> x = (12 m/s)² / (2 (0.6 m/s²)) = 120 m
c) Using the definition of average acceleration, slowing from 12 m/s to a stop in 6 s involves an acceleration of
(0 - 12 m/s) / (6 s) = -2 m/s²
