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Question
A particle moves along a line according to the law s(t) = 2t3 – 9t2 + 12t – 4, where t ≥ 0. At what times the particle changes direction?
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Solution
s = f(t) = 2t3 – 9t2 + 12t – 4
V = f'(t) = 6t2 – 18t + 12
V = 0 ⇒ 6(t2 – 3t – 2) = 0
(t – 1)(t – 2) = 0
t = 1, 2
When t < 1, (say t = 0.5)
V = 6(0.25 – 1.5 + 2) = +ve
When 1 < t < 2, (say t = 1.5)
V = 6(2.25 – 4.5 + 2) = – ve
When t > 2, (say t = 3)
V = 6(9 – 6 + 2) = +ve
So the particle changes its direction when t lies between 1 and 2 secs.
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