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)

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