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Asked in NEET 1994 · Acceleration by differentiation
Given: s=t³-6t²+3t+4.
Idea: locate the time at which the acceleration is zero, then read the velocity there.
v=(ds)/(dt)=3t²-12t+3.
a=(dv)/(dt)=6t-12.
Setting a=0: t=2 s.
v(2)=3(4)-12(2)+3=12-24+3=-9 m/s.
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