Physics › Motion In One Dimension › Rest and Motion
The relation between time ' $t$ ' and displacement ' $x$ ' is $t=\alpha x^2+\beta x$ where $\alpha$ and…
The relation between time ' $t$ ' and displacement ' $x$ ' is $t=\alpha x^2+\beta x$ where $\alpha$ and $\beta$ are constants. If ' $v$ ' is the velocity, the retardation is
$2 \alpha \nu \beta^2$ $2 \alpha \beta \nu^3$ $-2 \beta v^3$ $2 \alpha v^3$ Solution
$\begin{aligned}
\mathrm{t} & =\alpha \mathrm{x}^2+\beta \mathrm{x} \\
\therefore \frac{\mathrm{dt}}{\mathrm{dt}} & =2 \alpha \mathrm{x} \cdot \frac{\mathrm{dx}}{\mathrm{dt}}+\beta \cdot \frac{\mathrm{dx}}{\mathrm{dt}}=2 \alpha \mathrm{x} \cdot \mathrm{v}+\beta \cdot \mathrm{v} \\
\Rightarrow \mathrm{v} & =\frac{1}{2 \alpha \mathrm{x}+\beta}
\end{aligned}$
$\therefore \text { Retardation, } a=-\frac{\mathrm{dv}}{\mathrm{dt}}=-\frac{\mathrm{d}}{\mathrm{dt}}\left(\frac{1}{2 \alpha \mathrm{x}+\beta}\right)$
$\begin{aligned}
& =(2 \alpha x+\beta)^{-2} \cdot 2 \alpha \cdot v \\
& =v^2 \cdot 2 \alpha v=2 \alpha v^3
\end{aligned}$
Asked in: AP EAMCET 2024 (20 May Shift 2)
Practice more Motion In One Dimension questions on Aicharya