A cyclist accelerates at 0.6 m/s2 for 20 s. a) If he was not moving at the start, what speed does he reach?
b) How far does he travel in this 20 s period?
c) He brakes and stops in 6 s. Calculate his deceleration.

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

higher potential

explanation:

because it is in a high speed so it must move into a of high potential