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2.Motion in Straight Line
hard
સમય $'t'$ અને અંતર $'X'$ વરચે. સંબંધ $\mathrm{t}=\alpha \mathrm{x}^2+\beta \mathrm{x}$ છે. જ્યાં, $\alpha$ અને $\beta$ અચળાંકો છે. તો વેગ $(v)$ અને પ્રવેગ $(a)$ વરચે સંબંધ
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
