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 :
[b] 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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