A) \[\frac{\sqrt{3}\,{{v}_{0}}}{2\,\alpha }\]
B) \[{{\left( \frac{3\,{{v}_{0}}}{2\alpha } \right)}^{1/3}}\]
C) \[\sqrt{\frac{2\,{{v}_{0}}}{3\,\alpha }}\]
D) \[{{\left( \frac{3\,{{v}_{0}}^{2}}{2\alpha } \right)}^{1/3}}\]
Correct Answer: D
Solution :
Given, initial velocity \[={{V}_{0}}\] Final velocity =0 Deceleration, a = \[-\alpha {{x}^{2}}\] ...(i) Let the distance travelled by the particle be s. Now, we know that \[a=\frac{dv}{dt}=\frac{dv}{dt}\times \frac{dt}{dx}=\frac{vdv}{dx}\] Or \[a=v=\frac{dv}{dx}\] From Eq. (i) and (ii), we get \[v=\frac{dv}{dx}=-\alpha {{x}^{2}}\] Or \[vdx=-\alpha {{x}^{2}}dx\] On integrating with limit \[{{v}_{0}}\to 0\] and \[0\to s\] \[\int _{{{v}_{0}}}^{0}vdv=\int _{0}^{s}-\alpha {{x}^{2}}dx\] Or \[{{\left( \frac{{{v}^{2}}}{2} \right)}^{0}}n=-\alpha {{\left( \frac{{{x}^{3}}}{3} \right)}^{s}}\] \[\frac{-v_{0}^{2}}{2}=\frac{\alpha {{(s)}^{3}}}{3}\] \[\frac{v_{0}^{2}}{2}=\frac{\alpha {{s}^{3}}}{3}\] \[\frac{3v_{0}^{2}}{2\alpha }={{s}^{3}}\] \[s={{\left( \frac{3v_{0}^{2}}{2\alpha } \right)}^{1/3}}\]
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