Question

The distance x covered by a particle varies with time t as $x^2 = 2t^2 + 6t + 1$. Its acceleration varies with x as

• A. $x$
• B. $x^2$
• C. $x^{-1}$
• D. $x^{-3}$

Answer 1

Answer: (D)

Given : $x^{2} = 2t^{2} + 6t + 1\quad...\left(i\right)$
Differentiating with respect to t, we get
$2x \frac{dx}{dt} = 4t + 6$
$\therefore \quad x\upsilon = 2t + 3 \left(\because \frac{dx}{dt} = \upsilon\right)\quad...\left(ii\right)$
Again differentiating both sides w.r. to t we get,
$x \frac{d\upsilon}{dt} +\frac{\upsilon dx}{dt} = 2$
$xa + \upsilon^{2} = 2\quad\left(\because \frac{d\upsilon}{dt} =a \right)$
$a = \frac{2-\upsilon^{2}}{x}$
$a = \frac{1}{x}\left[2-\left[\frac{\left(2t+3\right)^{2}}{x}\right]\right]\quad$ Using $\left(ii\right)$
$= \frac{1}{x}\left[\frac{2x^{2}-\left(2t+3\right)^{2}}{x^{2} }\right]$
$ = \frac{1}{x} \left[\frac{2\left(2t^{2}+6t+1\right)-\left(4t^{2}+9+12t\right)}{x^{2}}\right]\quad$ Using $\left(i\right)$
$= \frac{1}{x} \left[\frac{4t^{2}+12rt+2-4t^{2}-9-12t}{x^{2}}\right]$ or $a \propto \frac{-1}{x^{3}}$