displacement and distance?

Question

Find the displacement and distance traveled by the moving object in the given interval.
a(t)=t^1/2 (ft/sec)/sec, v(0)=-18/ft/sec from t=0 to t=16 sec

Answer 1

Displacement:
a = t^(1/2)
v = v0 + (2/3)t^(3/2)
s = s0 + v0t + (4/15)t^(5/2)
s = 0 - (18)(16) + (4/15)(16)^(5/2)
s = (16)(- 18 + (4/15)(64))
s = (16)(- 14/15)
s ≈ - 14.933 ft.

Time at turn-around:
(2/3)t^(3/2) = 0 - v0
(2/3)t^(3/2) = 18
t^(3/2) = 27
t = 9 s
Distance to turn-around:
s = 0 - 18*9 + (4/15)(9)^(5/2)
s = 81(- 2 + (4/15)(9)^(1/2))
s = 81(- 2 + 4/5) = 81(- 1.2)
s = - 97.2
Distance traveled:
d = 97.2 + 97.2 - 14.933
d ≈ 179.4667 ft. ≈ 179.47 ft.