A) \[{{s}^{-\,3}}\]
B) \[{{s}^{3/2}}\]
C) \[{{s}^{-2/3}}\]
D) \[{{s}^{2}}\]
Correct Answer: A
Solution :
\[{{s}^{2}}=a{{t}^{2}}+2bt+c\text{ }\therefore \text{ 2s}\frac{ds}{dt}=2at+2b\] \[\text{or }\frac{ds}{dt}=\frac{at+b}{s},\text{ again differrentiating}\] \[\frac{{{d}^{2}}s}{d{{t}^{2}}}=\frac{a{{s}^{2}}-{{\left( at+b \right)}^{2}}}{{{s}^{3}}}\text{ }\therefore \text{a=}\frac{{{d}^{2}}s}{d{{t}^{2}}}\propto {{s}^{-3}}.\]
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