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The distance moved by the particle in time t is  given by  $s=t^{3}-12t^{2}+6t+8$ . At the instant , when its acceleration is zero the velocity is 

Answer:

Explanation:

$s=t^{3}-12t^{2}+6t+8$

$\frac{ds}{dt}=3t^{2}-24t+6$

$\frac{d^{2}s}{d^{2}t}=6t-24$

 Acceleration =0

 $\Rightarrow$ 6t-24=0

 $\therefore$   t=4

 Required velocity =  $\frac{ds}{dt}]_{t=4}$

$=3\times (4)^{2}-24\times 4+6$

 =48-96+6=42 units