A) \[28\,\,m/{{s}^{2}}\]
B) \[22\,\,m/{{s}^{2}}\]
C) \[12\,m/{{s}^{2}}\]
D) \[10\,\,m/{{s}^{2}}\]
Correct Answer: B
Solution :
\[v=4{{t}^{3}}-2t\] (given) \[\therefore \] \[a=\frac{dv}{dt}=12{{t}^{2}}-2\] and \[x=\int_{0}^{t}{v\ dt}=\int_{0}^{t}{(4{{t}^{3}}-2t)}\ dt={{t}^{4}}-{{t}^{2}}\] When particle is at 2m from the origin \[{{t}^{4}}-{{t}^{2}}=2\] Þ \[{{t}^{4}}-{{t}^{2}}-2=0\] \[({{t}^{2}}-2)\ ({{t}^{2}}+1)=0\] Þ \[t=\sqrt{2}\ \sec \] Acceleration at \[t=\sqrt{2\ }\ \sec \] given by, \[a=12{{t}^{2}}-2\]\[=12\times 2-2\]= \[22\ m/{{s}^{2}}\]
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