479 Horizontal Motion: The horizontal component of velocity ... - Wuala

91962_05_R1_p0479-0512 6/5/09 3:54 PM Page 502
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R1–33. The acceleration of a particle along a straight line is
defined by a = (2t - 9) m>s 2 , where t is in seconds. When
t = 0, s = 1 m and v = 10 m>s. When t = 9 s, determine
(a) the particle’s position, (b) the total distance traveled, and
(c) the velocity. Assume the positive direction is to the right.
a = (2t - 9)
dv = a dt
v
t
dv = (2t - 9) dt
L10
L 0
v - 10 = t 2 - 9t
v = t 2 - 9t + 10
ds = v dt
s
t
ds = At 2 - 9t + 10Bdt
L1
L 0
s - 1 = 1 3 t3 - 4.5t 2 + 10t
s = 1 3 t3 - 4.5t 2 + 10t + 1
Note v = 0 at t 2 - 9t + 10 = 0
At t = 1.298 s,
s = 7.127 m
At t = 7.702 s,
s = -36.627 m
At t = 9 s,
s = -30.50 m
t = 1.298 s and t = 7.702 s
a) s = -30.5 m
Ans.
b) s tot = (7.127 - 1) + 7.127 + 36.627 + (36.627 - 30.50) = 56.0 m Ans.
c) v| t = 9 = (9) 2 - 9(9) + 10 = 10 m>s
Ans.
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