A) \[18\,\,m/{{s}^{2}}\]
B) \[32\,\,m/{{s}^{2}}\]
C) \[29\,m/{{s}^{2}}\]
D) \[24\,\,m/{{s}^{2}}\]
Correct Answer: B
Solution :
The displacement of a particle along a straight line is \[s=3{{t}^{3}}+7{{t}^{2}}+14t+5....(i)\] Differentiating Eq. (i) with respect to time, which gives the velocity \[v=\frac{ds}{dt}\] \[=\frac{d}{dt}(3{{t}^{3}}+7{{t}^{2}}+14t+5)\] \[=\frac{d}{dt}(3{{t}^{3}})+\frac{d}{dt}(7{{t}^{2}})+\frac{d}{dt}(14t)+\frac{d}{dt}(5)\] \[v=3\frac{d}{dt}\,({{t}^{3}})+7\frac{d}{dt}({{t}^{2}})+14\frac{d}{dt}(t)+0....(ii)\] (as differentiation of a constant is zero) Now use \[\frac{d}{dt}({{x}^{n}})=n{{x}^{n-1}}\] \[So,v=3(3)\,{{t}^{3-1}}+7(2)\,({{t}^{2-1}})+14\,({{t}^{1-1}})\] \[\Rightarrow v=9{{t}^{2}}+14+14....(iii)\] \[(\because \,{{t}^{o}}=1)\] Again differentiating Eq. (iii) with respect to time, which gives the acceleration \[a=\frac{dv}{dt}=\frac{d}{dt}(9{{t}^{2}}+14t+14)\] \[=18t+14+0\] \[=18t+14\] At t = 1 s, a = 18 (1) + 14 \[=18+14\,=32\,m/{{s}^{2}}\]
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