Question

The relation between time and distance 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. $2{\beta }^2{v}^3$

Answer 1

Answer: (A)

$\frac{dt}{dx}=2\alpha x+\beta \Rightarrow v=\frac{1}{2\alpha x+\beta }$
$\because a=\frac{dv}{dt}=\frac{dv}{dx}\cdot \frac{dx}{dt}$
$a=v\frac{dv}{dx}=\frac{-v\cdot 2\alpha }{(2\alpha x+\beta {)}^2}=-2\alpha \cdot v\cdot v^2=-2\alpha v^3$
$\therefore$ Retardation $=2\alpha v^3$