A) \[28m{{s}^{-2}}\]
B) \[22m{{s}^{-2}}\]
C) \[12m{{s}^{-2}}\]
D) \[10m{{s}^{-2}}\]
Correct Answer: B
Solution :
Velocity \[v=4{{t}^{3}}-2t\] ... (i) \[\frac{dx}{dt}=4\,\,{{t}^{3}}-2t\] On integration, we get \[x=2={{t}^{4}}-{{t}^{2}}={{\alpha }^{2}}-\alpha \] ? (ii) \[(if\,\,{{t}^{2}}=\alpha )\] \[{{\alpha }^{2}}-\alpha -2=0\] \[(\alpha -2)(\alpha +1)=0\] \[\alpha =2\] \[\alpha =-1\], which is not possible \[{{t}^{2}}=\alpha =2\]or\[t=\sqrt{2}\] Differentiating Eq. (i), w.r.t.\[t\] \[\frac{dv}{dt}=12{{t}^{2}}-2\] \[\alpha =12\times 2-2=22m{{s}^{-2}}\]
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