Relation between time ' t ' and distance ' x ' is t= α x^2+β x , where α and β a

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2.Motion in Straight Line

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The relation between time ' $t$ ' and distance ' $x$ ' is $t=$ $\alpha x^2+\beta x$, where $\alpha$ and $\beta$ are constants. The relation between acceleration $(a)$ and velocity $(v)$ is:

A

$a=-2 \alpha v^3$

B

$a=-5 \alpha v^5$

C

$a=-3 \alpha v^2$

D

$a=-4 \alpha v^4$

(JEE MAIN-2024)

Solution

$\mathrm{t}=\alpha \mathrm{x}^2+\beta \mathrm{x} \text { (differentiating wrt time) }$

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

$\frac{1}{\mathrm{v}}=2 \alpha \mathrm{x}+\beta$

$\text { (differentiating wrt time) }$

$-\frac{1}{\mathrm{v}^2} \frac{\mathrm{dv}}{\mathrm{dt}}=2 \alpha \frac{\mathrm{dx}}{\mathrm{dt}}$

$\frac{\mathrm{dv}}{\mathrm{dt}}=-2 \alpha \mathrm{v}^3$

Standard 11

Physics

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