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