A) \[zero\]
B) \[2\,\,m\]
C) \[3\,\,m\]
D) \[-2\,\,m\]
Correct Answer: A
Solution :
Key Idea: Acceleration is equal to rate of change of velocity. Given, \[s={{t}^{3}}-3{{t}^{2}}+2\] Velocity \[v=\frac{ds}{dt}=3{{t}^{2}}-6t\] Acceleration \[a=\frac{{{d}^{2}}s}{d{{t}^{2}}}=\frac{dv}{dt}=6t-6\] At\[a=0\], we have \[6t-6=0\] \[\Rightarrow \] \[t=1s\] Hence, \[s={{(1)}^{3}}-3{{(1)}^{2}}+2\] \[=1-3+2=0\]You need to login to perform this action.
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