A) \[1/{{s}^{3}}\]
B) \[1/s\]
C) \[~1/{{s}^{4}}\]
D) \[1/{{s}^{2}}\]
Correct Answer: A
Solution :
Given, \[s=\sqrt{a{{t}^{2}}+2bt+c}\] \[\Rightarrow \] \[\frac{ds}{dt}=\frac{2at+2b}{2\sqrt{a{{t}^{2}}+2bt+c}}=\frac{at+b}{\sqrt{a{{t}^{2}}+2bt+c}}\] \[\Rightarrow \] \[\frac{{{d}^{2}}s}{d{{t}^{2}}}=\frac{\left[ \begin{align} & \sqrt{a{{t}^{2}}+2bt+c}\times a\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \\ & -(at+b)\frac{(2at+2b)}{2\sqrt{a{{t}^{2}}+2bt+c}} \\ \end{align} \right]}{{{(\sqrt{a{{t}^{2}}+2bt+c})}^{3}}}\] \[=\frac{{{a}^{2}}{{t}^{2}}+2abt+ac-({{a}^{2}}{{t}^{2}}+{{b}^{2}}+2abt)}{{{(\sqrt{a{{t}^{2}}+2bt+c})}^{3}}}\] \[\Rightarrow \] \[\frac{{{d}^{2}}s}{d{{t}^{2}}}=\frac{ac-{{b}^{2}}}{{{s}^{3}}}\] \[\Rightarrow \] Acceleration \[\propto \frac{1}{{{s}^{3}}}\]
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