### 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

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