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Question
A particle moves along a line according to the law s(t) = 2t3 – 9t2 + 12t – 4, where t ≥ 0. Find the particle’s acceleration each time the velocity is zero
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Solution
s(t) = 2t3 – 9t2 + 12t – 4
Velocity v = `"ds"/"dt"`
= 6t2 – 18t + 12
v = 0
⇒ 6(t2 – 3t + 2) = 0
⇒ t = 1, 2
Acceleration = `("d"^2"s")/("dt"^2)`
= 12t – 18
At t = 1, Acceleration = 12(1) – 18 = – 6 m/sec2
At t = 2, Acceleration = 12 (2) – 18 = 6 m/sec2
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