Chapter 8 Question 13 c

c.) 0 = 3t2 – 6t
t = 2,0
test t = 1, v(1) = 3(1)2 – 6(1)
v(1) = -3
test t = 3, v(3) = 3(3)2 – 6(3)
v(3) = 9
therefore, 0 <> 2 is positive
0 - (3t2 – 6t) dt
[-t3 + 3t2]
[(-(2)3 + 3(2)2) – 0]
= 4
3t2 - 6t dt
[t3 – 3t2]
[((4)3 – 3(4)2) – ((2)3 – 3(2)2)]
[16 – (-4)]
= 20
total distance traveled = 20 + 4
= 24 units