A) \[\frac{\text{m}}{\text{3P}}\text{(v}\,_{\text{2}}^{\text{3}}\text{-v}\,_{\text{1}}^{\text{3}}\text{)}\]
B) \[\frac{\text{m}}{\text{3P}}\text{(}{{\text{v}}_{2}}\text{-}{{\text{v}}_{1}}\text{)}\]
C) \[\frac{3P}{m}\text{(v}\,_{2}^{2}\text{-v}\,_{1}^{2}\text{)}\]
D) \[\frac{m}{3P}\text{(v}\,_{2}^{2}\text{-v}\,_{1}^{2}\text{)}\]
Correct Answer: A
Solution :
\[P=Fv=mav\] \[a=\frac{P}{mv}\] or \[v\frac{dv}{ds}=\frac{P}{mv}\] or \[{{v}^{2}}dv=\frac{P}{m}ds\] or \[\frac{P}{m}\int_{0}^{s}{ds=\int_{{{v}_{1}}}^{{{v}_{2}}}{{{v}^{2}}dv}}\] or \[\frac{P}{m}s=\frac{1}{3}(v_{2}^{3}-v_{1}^{3})\] \[s=\frac{m}{3P}(v_{2}^{3}-v_{1}^{3})\]
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