Question

A particle moves a distance x in time t according to equation $x^2=1+t^2$ The acceleration of the particle is

• 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)

Given, $x^2=t^2+1\dots$(i)
Differentiating with respect to $t$ on both sides, we get
$2x\frac{dx}{dt}=2t$ or $xv=t\dots$ (ii)
Again differentiating w.r.t. $t$ on both sides, we get
$x\frac{dv}{dt}+v\frac{dx}{dt}=1$ or
$x\frac{dv}{dt}=1-v^2$
or $\frac{dv}{dt}=\frac{1-v^2}{x}=\frac{1-\frac{t^2}{x^2}}{x}$
$=\frac{x^2-t^2}{x^3}=\frac{1}{x^3}$ (Using (i, ii))
$\therefore$ Acceleration, $a=\frac{dv}{dt}=\frac{1}{x^3}$