A) -9
B) -12
C) 42
D) 3
Correct Answer: A
Solution :
Given \[s={{t}^{3}}-6{{t}^{2}}+3t+4\] \[\frac{ds}{dt}=3{{t}^{2}}-12t+3\] and \[\frac{{{d}^{2}}s}{d{{t}^{2}}}=6t-12\] If acceleration is zero, \[6t-12=0\] or \[t=2\sec \] Hence, velocity at t = 2 sec \[v=\frac{ds}{dt}=3{{t}^{2}}-12t+3\] \[=12-24+3=-\,9\,m\text{/}s\]You need to login to perform this action.
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