A) \[\frac{1}{{{s}^{2}}}\]
B) \[\frac{1}{s}\]
C) \[\frac{1}{{{s}^{3}}}\]
D) \[{{s}^{3}}\]
E) \[{{s}^{2}}\]
Correct Answer: C
Solution :
Given that, \[s=\sqrt{a{{t}^{2}}+2bt+c}\] On differentiating w.r.t. t, we get \[\frac{ds}{dt}=\frac{1}{2\sqrt{a{{t}^{2}}+2bt+c}}(2at+2b)\] \[\Rightarrow \] \[\frac{ds}{dt}=\frac{at+b}{\sqrt{a{{t}^{2}}+2bt+c}}=\frac{at+b}{s}\] Again differentiating, we get \[\frac{{{d}^{2}}s}{d{{t}^{2}}}=\frac{sa-(at+b)\frac{ds}{dt}}{{{s}^{2}}}\] \[=\frac{sa-(at+b)\left( \frac{at+b}{s} \right)}{{{s}^{2}}}\] \[=\frac{{{s}^{2}}a-{{(at+b)}^{2}}}{{{s}^{3}}}\] \[=\frac{(a{{t}^{2}}+2bt+c)a-{{(at+b)}^{2}}}{{{s}^{3}}}\] \[=\frac{ac-{{b}^{2}}}{{{s}^{3}}}\] \[\Rightarrow \]Acceleration varies as\[\frac{1}{{{s}^{3}}}\].
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