Question

A particle moves along a straight line such that its displacement at any time $t$ is given by $S = t^{3} - 6t^{2} + 3t + 4$ metres The velocity when the acceleration is zero is

• A. $3ms^{- 1}$
• B. $- 12ms^{- 1}$
• C. $42ms^{- 1}$
• D. $- 9ms^{- 1}$

Answer 1

Answer: (D)

$- 9ms^{- 1}$

Answer 2

$v = \frac{ds}{dt} = 3t^{2} - 12t + 3$ and $a = \frac{dv}{dt} = 6t - 12$

For $a = 0$, we have $t = 2$ and at $t = 2,\mspace{6mu} v = - 9\mspace{6mu} ms^{- 1}$