Question

A particle moves rectilinearly. Its displacement $x$ at time $t$ is given by $x^2 = at^2 + b$ where $a$ and $b$ are constants. Its acceleration at time $t$ is proportional to

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

Answer 1

Answer: (A)

$\frac{1}{x^{3}}$

Answer 2

Given : $x^{2}=at^{2}+b\quad\ldots\left(i\right)$ Differentiating w.r.t. $t$ on both sides, we get $2x \frac{dx}{dt}=2at$ ; $xv=at\quad\left(\because v=\frac{dx}{dt}\right)$ Again differentiating w.r.t. $t$ on both sides, we get $x \frac{dv}{dt}+v \frac{dx}{dt}=a$ or $x \frac{dv}{dt}=a-v^{2}$ $\frac{dv}{dt}=\frac{a-v^{2}}{x}=\frac{a-\left(\frac{at}{x}\right)^{2}}{x}$ $=\frac{a-\frac{a^{2}t^{2}}{x^{2}}}{x}=\frac{a\left(x^{2}-at^{2}\right)}{x^{3}}$ $\frac{dv}{dt}=\frac{ab}{x^{3}}$ or Acceleration $\propto \frac{1}{x^{3}}\quad$ (Using $(i)$)