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