A) 1 sec
B) 2 sec
C) 1, 2 sec
D) None of these
Correct Answer: C
Solution :
\[\frac{ds}{dt}=6{{t}^{2}}-18t+12\] = velocity = 0 (when particle stopped) Þ \[6{{t}^{2}}-18t+12=0\Rightarrow (t-1)\,(t-2)=0\] Hence time 1, 2 sec.You need to login to perform this action.
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