If v = x^2 -5x + 4 , acceleration of particle when velocity of particle is zero

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2.Motion in Straight Line

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If $v = x^2 -5x + 4$, find the acceleration of particle when velocity of the particle is zero

A

$4$

B

$1$

C

$3$

D

$0$

Solution

$a=\frac{d v}{d t}=\frac{d v}{d x} \cdot \frac{d x}{d t}=v \frac{d v}{d x}$

$\therefore a=v \frac{d v}{d x}$

so when $\mathrm{v}=0$ we get $a=0$

$O R$

$v=x^{2}-5 x+4$

$\therefore \frac{\mathrm{dv}}{\mathrm{dt}}=2 \mathrm{x} \frac{\mathrm{dx}}{\mathrm{dt}}-5 \frac{\mathrm{dx}}{\mathrm{dt}}+0$

$\Rightarrow$ acc. $^{\prime} a^{\prime}=(2 x-5) v$

$\text { when } \mathrm{v}=0, \mathrm{a}=0$

Standard 11

Physics

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