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

Answer 1

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