A) \[\frac{1}{{{x}^{3}}}\]
B) \[\frac{1}{x}\,-\frac{1}{{{x}^{2}}}\]
C) \[-\frac{t}{{{x}^{2}}}\]
D) \[\frac{1}{x}\,-\,\frac{{{t}^{2}}}{{{x}^{3}}}\]
Correct Answer: D
Solution :
| [d] \[{{x}^{2}}=(1+{{t}^{2}})\] or \[x={{(1\text{ }+\text{ }{{t}^{2}})}^{1/2}}\] |
| Velocity, \[\frac{dx}{dt}=\,\frac{1}{2}\,{{(1+\,{{t}^{2}})}^{-1/2}}\times \,2t\] |
| \[=t{{(1+\,{{t}^{2}})}^{-1/2}}\] |
| Acceleration, \[\frac{{{d}^{2}}x}{d{{t}^{2}}}\,=t\left( -\frac{1}{2} \right)\,\times \,{{\left( 1+{{t}^{2}} \right)}^{-3/2}}\] |
| \[\times \,2t+{{(1+{{t}^{2}})}^{-1/2}}\] |
| \[=\,\frac{1}{x}-\frac{{{t}^{2}}}{{{x}^{3}}}\] |
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