A particle moves along a straight line such that its displacement at any time $t$ is given by $s = (t^{3} - 6 t^{2} + 3 t + 4) m$ . Find the velocity when the acceleration is zero.

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Solution

$s = t^{3} - 6 t^{2} + 3 t + 4 \Rightarrow v = 3 t^{2} - 12 t + 3$
$a = d v / d t = 6 t - 12 = 0 \Rightarrow t = 2 s$
$v = 3 (2)^{2} - 12 \times 2 + 3 = - 9 m s^{- 1}$

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A particle moves along a straight line such that its displacement at any time t is given by s=(t3−6t2+3t+4)m. Find the velocity when the acceleration is zero.

s=t3−6t2+3t+4⇒v=3t2−12t+3a=dv/dt=6t−12=0⇒t=2sv=3(2)2−12×2+3=−9ms−1

A particle moves along a straight line such that its displacement at any time $t$ is given by $s = (t^{3} - 6 t^{2} + 3 t + 4) m$ . Find the velocity when the acceleration is zero.

$s = t^{3} - 6 t^{2} + 3 t + 4 \Rightarrow v = 3 t^{2} - 12 t + 3$
$a = d v / d t = 6 t - 12 = 0 \Rightarrow t = 2 s$
$v = 3 (2)^{2} - 12 \times 2 + 3 = - 9 m s^{- 1}$