- JEE Mains 2020
- Hard
- Numerical
The distance \(x\) covered by a particle in one dimensional motion varies with time \(t\) as \(\mathrm{x}^{2}=\mathrm{at}^{2}+2 \mathrm{bt}+\mathrm{c.}\) If the acceleration of the particle depends on \(\mathrm{x}\) as \(\mathrm{x}^{-\mathrm{n}},\) where \(\mathrm{n}\) is an integer, the value of \(\mathrm{n}\) is
Answer & Solution
(D) \(3\)
Step-by-step Solution
Detailed explanation
Differentiating w.r.t. time
\(\frac{d x}{d t}=v=\frac{1}{2 \sqrt{a t^{2}+2 b t+c}} \times(2 a t+2 b)\)
\(\Rightarrow \mathrm{v}=\frac{\mathrm{at}+\mathrm{b}}{\mathrm{x}}\)
\(\Rightarrow \mathrm{vx}=\mathrm{at}+\mathrm{b}\)
Differentiating w.r.t. \(\mathrm{x}\)
\(\Rightarrow \frac{d v}{d x} \times x+v=a \times \frac{d t}{d x}\)
Multiply both side by v
\(\Rightarrow\left(v \frac{d v}{d x}\right) x+v^{2}=a\)
\(\Rightarrow \mathrm{a}^{\prime} \mathrm{x}=\mathrm{a}-\mathrm{v}^{2}\) [Here \(a'\) is acceleration]
\(\Rightarrow \mathrm{a}^{\prime} \mathrm{x}=\mathrm{a}-\left(\frac{\mathrm{at}+\mathrm{b}}{\mathrm{x}}\right)^{2}\)
\(\Rightarrow \mathrm{a}^{\prime} \mathrm{x}=\frac{\mathrm{ax}^{2}-(\mathrm{at}+\mathrm{b})^{2}}{\mathrm{x}^{2}}\)
\(\Rightarrow \mathrm{a}^{\prime} \mathrm{x}=\frac{\mathrm{a}\left(\mathrm{at}^{2}+2 \mathrm{bt}+\mathrm{c}\right)-(\mathrm{at}+\mathrm{b})^{2}}{\mathrm{x}^{2}}\)
\(\Rightarrow \mathrm{a}^{\prime} \mathrm{x}=\frac{\mathrm{ac}-\mathrm{b}^{2}}{\mathrm{x}^{2}}\)
\(\Rightarrow \mathrm{a}^{\prime}=\frac{\mathrm{ac}-\mathrm{b}^{2}}{\mathrm{x}^{3}}\)
\(\therefore a^{\prime} \propto \frac{1}{x^{3}} \quad \therefore n=3\)
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