1) average velocity = (v(8) - v(0))/(8 - 0) = (10 - 18)/8 = -1m/s

2) v(5) = 5^2 - 9(5) + 18 = 25 - 45 + 18 = -2m/s

3) set v(t) = 0

t^2 - 9t + 18 = 0

(t - 6)(t - 3) = 0

t = 6 t = 3

When 0
4) find the acceleration function

a(t) = 2t - 9

2t - 9 = 0

2t = 9

t = 9/2

When 0
5) Int(t^2 -9t + 18) = (t^3)/3 - (9t^2)/2 + 18t + c

s(0) = 1

s(t) = (t^3)/3 - (9t^2)/2 + 18t + 1

Now we use the 3 intervals in part 3 to find the total distance

s(3) - s(0) = (9 - 81/2 + 54 + 1) - 1 = 45/2
s(6) - s(3) = ( 72 -162 + 108 + 1) - 47/2 = -9/2
s(8) - s(6) = (512/3 - 288 + 144 + 1) - 19 = 26/3

All distances are scalar quantities so they are positive

45/2 + 9/2 + 26/3 = 107/3 meters


IS THIS CORRECT?