Question

The relation between time $t$ and distance $x$ is $t=\alpha x^2+\beta x$ where $\alpha$ and $\beta$ are constants. The retardation is

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

Answer 1

The correct option is A

$-2\alpha v^3$

Explanation for the correct answer:

Compute the retardation:

$t=\alpha x^2+\beta x$

Differentiate with respect to $t$

$2\alpha x\frac{dx}{dt}+\beta \frac{dx}{dt}=1$

$\Rightarrow$ $\frac{dx}{dt}=\frac{1}{\left(2\alpha x+\beta \right)}$

$\Rightarrow$ $v=\frac{1}{\left(2\alpha x+\beta \right)}$

$a=\frac{dv}{dt}=\frac{dv}{dt}\times \frac{dx}{dx}$

$=\frac{dv}{dx}\times \frac{dx}{dt}$

$\Rightarrow$ $a=v\frac{dv}{dx}$

$\frac{dv}{dx}=\frac{d}{dx}{\left(2\alpha x+\beta \right)}^{-1}$

$=\frac{-2\alpha }{{\left(2\alpha x+\beta \right)}^2}$

$=-2\alpha v^2$

$\therefore$ $a=v\times \left(-2\alpha v^2\right)$

$\therefore$ $a=-2\alpha v^3$

Hence, the retardation is $-2\alpha v^3$.

Hence, option $\left(A\right)$ is the correct answer.