A) \[\frac{4a{{t}^{3}}}{2}\]
B) \[\left( \frac{\sqrt{\alpha }}{6} \right)t\]
C) \[\left( \frac{{{\alpha }^{2}}}{36} \right){{t}^{4}}\]
D) \[\left( \frac{{{\alpha }^{2}}}{144} \right){{t}^{4}}\]
Correct Answer: D
Solution :
| [d] \[a=\alpha \sqrt{x}\] |
| \[\Rightarrow \] \[\frac{vdv}{dx}=\alpha \sqrt{x}\] \[\Rightarrow \] \[vdv=\alpha \sqrt{x}dx\] |
| \[\int_{0}^{v}{vdv}=\int_{0}^{x}{\alpha \sqrt{x}}dx\] |
| \[\Rightarrow \] \[\frac{{{v}^{2}}}{2}=\frac{(\alpha {{x}^{3/2}})}{(3/2)}\] |
| \[\Rightarrow \] \[{{v}^{2}}=\frac{4\alpha {{x}^{3/2}}}{3}\] \[\Rightarrow \] \[v=\left( \sqrt{\frac{4\alpha }{3}} \right){{x}^{3/4}}\] |
| \[\Rightarrow \] \[\frac{dx}{dt}=\left( \sqrt{\frac{4\alpha }{3}} \right){{x}^{3/4}}\] \[\Rightarrow \] \[\frac{dx}{{{x}^{3/4}}}=\left( \sqrt{\frac{4\alpha }{3}} \right)dt\] |
| \[\Rightarrow \] \[4{{x}^{1/4}}=\left( \sqrt{\frac{4\alpha }{3}} \right)t\] \[\Rightarrow \] \[x=\frac{{{\alpha }^{2}}{{t}^{4}}}{144}\] |
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