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 \[t=1\,s,\] |
| \[a=18\text{ (1)}+14\] |
| \[=18+14=32\text{ }m/{{s}^{2}}\] |
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