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  • The distance  covered by a particle in one dimensional motion varies with time t as <img alt="x^2 = at^2+2bt + c" src="/latex-image/?x%5E2%20%3D%20at%5E2&plus;2bt%

The distance x covered by a particle in one dimensional motion varies with time t as x^2 = at^2+2bt + c. If the acceleration of the particle depends of x as x^{-n}, where n is an integer, the value of n is ______
Option: 1 3
Option: 2 4
Option: 3 1
Option: 4  0
 

Answers (1)

best_answer

 

x^2 = at^2+2bt + c \ \ \ - \left ( 1 \right )

\text{Differentiating equation 1 with respect to 't', we get }

2x\frac{\text d x}{\text dt} = 2at + 2b \ \ \ \ -\left ( 2 \right )
\text{Again differentiating equation 2 with respect to 't', we get }


\begin{gathered} 2\left[x \frac{d^{2} x}{d t^{2}}+\frac{d x}{d t} \cdot \frac{d x}{d t}\right]=2 a \\ x \frac{d^{2} x}{d t^{2}}+\left(\frac{dx}{d t}\right)^{2}=\frac{2 a}{a} \end{gathered} \\ \text{Put the value of}\ \frac{d x}{d t}\ \text{from eq.} (2) \text{ we have} \\ x \frac{d^{2} x}{d t^{2}}+\left(\frac{2 a t+2 b}{2 x}\right)^{2}=2\\ \\ \frac{d^{2} x}{d t^{2}}=-\frac{(2 at+2 b)^{2}}{4 x^{2} x}=\frac{-(2 at+2 b)^{2}}{4 x^{3}}\\ \\ or \frac{d^{2} x}{d x^{2}} \propto \frac{A}{x^{3}}\left\{\begin{array}{l}A \text { is constant here i.e }A=\frac{(-2at+2b)^{2}}{4} \\ \\ \end{array}\right\}\\ \\\frac{d^{2}x}{dt^{2}} \propto x^{-3}

\therefore n=3

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