Question

The displacement $x$ of a particle at time $t$ moving along a straight line path is given by $x^2=a{t}^2+2bt+c,$ where $a,b$ and $c$ are constants. The acceleration of the particle varies as

• A. $x^{-1}$
• B. $x^{-2}$
• C. $x^{-3}$
• D. $x^{-4}$

Answer 1

Answer: (C)

$x^2=a{t}^2+2bt+c\dots$(i)
Differentiating equation (i) w.r.t. time $t$, we get
$2x\frac{dx}{dt}=2at+2b$ or
$x\frac{dx}{dt}=at+b$
or $xv=at+b$
$(\because$ Velocity, $v=\frac{dx}{dt})\dots$(ii)
Differentiating equation (ii) w.r.t. time $t$, we get
$x\frac{dv}{dt}+v\frac{dx}{dt}=a$ or $x\frac{dv}{dt}=a-v^2$
Acceleration $=\frac{dv}{dt}=\frac{a-v^2}{x}$
$=\frac{a-{\left(\frac{at+b}{x}\right)}^2}{x}=\frac{a{x}^2-(at+b{)}^2}{x^3}$
(Using (ii))
$=\frac{a\left(a{t}^2+2bt+c\right)-(at+b{)}^2}{x^3}=\frac{ac-b^2}{x^3}$ (Using(i))