A) \[{{t}^{2}}\]
B) \[t\]
C) \[{{t}^{1/2}}\]
D) \[{{t}^{3}}\]
Correct Answer: A
Solution :
| [a] The variation of the velocity is given as. |
| \[v=\alpha \sqrt{x}\] |
| We can write it as \[\frac{dx}{dt}=\alpha \sqrt{x}\] \[\left( \because v=\frac{dx}{dt} \right)\] |
| \[\Rightarrow \] \[\frac{dx}{\sqrt{x}}=\alpha dt\] |
| Perform integration within the limit \[\int_{0}^{x}{\frac{dx}{\sqrt{x}}}=\int_{0}^{t}{\alpha \,dt}\] |
| [\[\because \]at\[t=0,\text{ }x=0\]and let at any time t, particle be at\[x\]] |
| \[\Rightarrow \] \[\left. \frac{{{x}^{1/2}}}{1/2} \right|_{0}^{x}=\alpha t\] |
| \[\Rightarrow \] \[{{x}^{1/2}}=\frac{\alpha }{2}t\] |
| \[\Rightarrow \] \[x=\frac{{{\alpha }^{2}}}{4}\times {{t}^{2}}\]\[\Rightarrow \]\[x\propto {{t}^{2}}\] |
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