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 :
\[x=\alpha {{t}^{3}}\] and \[y=\beta {{t}^{3}}\] \[{{v}_{x}}=\frac{dx}{dt}=3\alpha t\] and \[{{v}_{y}}=\frac{dy}{dt}=3\beta {{t}^{2}}\] \[\therefore \,\,\,v=\sqrt{v_{x}^{2}+v_{y}^{2}}=\sqrt{9{{\alpha }^{2}}{{t}^{4}}+9{{\beta }^{2}}{{t}^{4}}}=3{{t}^{2}}\sqrt{{{\alpha }^{2}}+{{\beta }^{2}}}\]
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