Question

The displacement $x$ of a particle at time $t$ moving along a straight line path is given by $x^{2}=a t^{2}+2 b t+c,$ 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}+2 b t+c \dots $(i)
Differentiating equation (i) w.r.t. time $t$, we get
$2 x \frac{d x}{d t}=2 a t+2 b$ or
$ x \frac{d x}{d t}=a t+b$
or $x v=a t+b$
$\left(\because\right.$ Velocity, $\left.v=\frac{d x}{d t}\right) \dots$(ii)
Differentiating equation (ii) w.r.t. time $t$, we get
$x \frac{d v}{d t}+v \frac{d x}{d t}=a $ or $ x \frac{d v}{d t}=a-v^{2}$
Acceleration $=\frac{d v}{d t}=\frac{a-v^{2}}{x} $
$=\frac{a-\left(\frac{a t+b}{x}\right)^{2}}{x}=\frac{a x^{2}-(a t+b)^{2}}{x^{3}}$
(Using (ii))
$=\frac{a\left(a t^{2}+2 b t+c\right)-(a t+b)^{2}}{x^{3}}=\frac{a c-b^{2}}{x^{3}}$ (Using(i))