A) \[3t\,\sqrt{{{\alpha }^{2}}+{{\beta }^{2}}}\]
B) \[3{{t}^{2}}\,\sqrt{{{\alpha }^{2}}+{{\beta }^{2}}}\]
C) \[{{t}^{2}}\,\sqrt{{{\alpha }^{2}}+{{\beta }^{2}}}\]
D) \[\sqrt{{{\alpha }^{2}}+{{\beta }^{2}}}\]
Correct Answer: B
Solution :
The positions of particle is given \[x=\alpha {{t}^{3}},y=\beta {{t}^{3}}\] On differentiating with respect to t, we get \[{{v}_{x}}=\frac{dx}{dt}=3\alpha {{t}^{2}},{{v}_{y}}=\frac{dy}{dt}=3\beta {{t}^{2}}\] Resultant velocity \[v=\sqrt{{{v}_{x}}^{2}+{{v}_{y}}^{2}}\] \[=\sqrt{9{{\alpha }^{2}}{{t}^{4}}+9{{\beta }^{2}}{{t}^{4}}}\] \[=3{{t}^{2}}\sqrt{{{\alpha }^{2}}+{{\beta }^{2}}}\]
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