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

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Solution

$s = t^{3} - 6 t^{2} + 3 t + 4. v = \frac{d s}{d t} = 3 t^{2} - 12 t + 3$

$a = \frac{d v}{d t} = 6 t - 12$

Put $a = 0 \Rightarrow 6 t - 12 = 0 \Rightarrow t = 2 s$

$v_{t = 2 s} = 3 (2)^{2} - 12 \times 2 + 3 = - 9 m s^{- 1}$

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

s=t3−6t2+3t+4.v=dsdt=3t2−12t+3a=dvdt=6t−12Put a=0⇒6t−12=0⇒t=2svt=2s=3(2)2−12×2+3=−9ms−1

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

$s = t^{3} - 6 t^{2} + 3 t + 4. v = \frac{d s}{d t} = 3 t^{2} - 12 t + 3$

$a = \frac{d v}{d t} = 6 t - 12$

Put $a = 0 \Rightarrow 6 t - 12 = 0 \Rightarrow t = 2 s$

$v_{t = 2 s} = 3 (2)^{2} - 12 \times 2 + 3 = - 9 m s^{- 1}$