Question

The relation between time $ t $ and distance $ x $ for a moving particle is $ t = \alpha x^{2} + \beta x $ , where $ \alpha $ and $ \beta $ are constants. If $ v $ is the velocity at distance $ x $ , then the retardation of the particle is

• A. $ 2\alpha v^{3} $
• B. $ 2\beta v^{3} $
• C. $ 2\alpha\beta v^{3} $
• D. $ 2 \beta^{2}v^{2} $

Answer 1

Answer: (A)

Given that,
$t=\alpha x^{2}+\beta x$
Differentiating with respect to '$t$'
$1=\alpha\cdot\frac{2xdx}{dt}+\beta \frac{dx}{dt}$
$1=2\alpha xv+\beta v$
$1=v\left(2\alpha x+\beta\right)$
$v=\frac{1}{2\alpha x+\beta} \ldots\left(i\right)$
Retardation, $a=-\frac{dv}{dt}$
$=-\frac{d}{dt} \left(\frac{1}{2\alpha x+\beta}\right)$
$=\left(-\frac{1}{\left(2\alpha x+\beta\right)^{2}}\right)\cdot\left(-2\alpha\frac{dx}{dt}\right)$
$=\frac{1}{\left(2\alpha x+\beta\right)^{2}}\cdot2\alpha v$
From Eq.$ \left(i\right)$, we get
$=v^{2}\cdot2\alpha v $
$=2\alpha v^{3}$