A) \[\pi \]
B) \[(2a{{x}^{2}}x+6b{{y}^{2}})\,\,\hat{k}\]
C) \[(4b{{c}^{2}}x+3a{{c}^{2}})\,\,\hat{k}\]
D) \[(4{{c}^{2}}x+2b{{y}^{2}})\,\,\hat{k}\]
Correct Answer: A
Solution :
[A]| Given that \[\frac{dx}{dt}=\frac{dy}{dt}=c\] |
| \[\therefore \frac{{{d}^{2}}x}{d{{t}^{2}}}\] |
| \[\therefore \frac{{{d}^{2}}x}{d{{t}^{2}}}=\frac{{{d}^{2}}y}{d{{t}^{2}}}=0\] |
| Further \[z=a{{x}^{3}}+b{{y}^{2}}\] |
| \[\therefore \frac{dz}{dt}=3\,a{{x}^{2}}\frac{dx}{dt}+2\,by\frac{dy}{dt}\]\[=3\,ac{{x}^{2}}+2\,bcy\left( \frac{dx}{dt}=c=\frac{dt}{dt} \right)\] |
| \[\therefore \frac{{{d}^{2}}x}{d{{t}^{2}}}=6\,acx\left( \frac{dx}{dt} \right)+2\,bc\left( \frac{dy}{dt} \right)\]\[=6a{{c}^{2}}x+2b{{c}^{2}}\] |
| Now acceleration of particle is \[\vec{a}=\frac{{{d}^{2}}x}{d{{t}^{2}}}\,\,\hat{i}+\frac{{{d}^{2}}x}{d{{t}^{2}}}\,\,\hat{j}+\frac{{{d}^{2}}x}{d{{t}^{2}}}\,\,\hat{k}\]\[=(6a{{c}^{2}}x+2b{{c}^{2}})\,\,\hat{k}\] |
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