A particle is moving with velocity v (t) = t2 _ 9t + 18 with distance, s measured in meters, left or right of zero, and t measured in seconds, with t between 0 and 8 seconds inclusive. The position at time t = 0 sec is 1 meter right of zero, that is, s (0) = 1.

The average velocity over the interval 0 to 8 seconds

The instantaneous velocity and speed at time 5 secs

The time interval (s) when the particle is moving right

The time interval (s) when the particle is

going faster

slowing down

Find the total distance the particle has traveled between 0 and 8 seconds

V = t^2 - 9t + 18

position, s

s = t^3 / 3 - 4.5t^2 + 18t + C

t = 0, s = 1 = > 1=C = > s = t^3/3 - 4.5t^2 + 18t + 1

Average velocity: distance / time

distance: t = 8 = > s = 8^3 / 3 - 4.5 (8) ^2 + 18 (8) + 1 = 27.67 m

Average velocity = 27.67 / 8 = 3.46 m/s

t = 5 s

v = t^2 - 9t + 18 = 5^2 - 9 (5) + 18 = - 2 m/s

speed = |-2| m/s = 2 m/s

Moving right

V > 0 = > t^2 - 9t + 18 > 0

(t - 6) (t - 3) > 0

=> t > 6 and t > 3 = > t > 6 s = > Interval (6,8)

=> t < 6 and t t interval (0,3)

Going faster and slowing dowm

acceleration, a = v' = 2t - 9

a > 0 = > 2t - 9 > 0 = > 2t > 9 = > t > 4.5 s

Then, going faster in the interval (4.5, 8) and slowing down in (0, 4.5)