MCQ
If $v = x^2 -5x + 4$, find the acceleration of particle when velocity of the particle is zero
- A$4$
- B$1$
- C$3$
- ✓$0$
✓
Answer
Correct option: D.
$0$
d
$a=\frac{d v}{d t}=\frac{d v}{d x} \cdot \frac{d x}{d t}=v \frac{d v}{d x}$
$a=\frac{d v}{d t}=\frac{d v}{d x} \cdot \frac{d x}{d t}=v \frac{d v}{d x}$
$\therefore a=v \frac{d v}{d x}$
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. $^{\prime} a^{\prime}=(2 x-5) v$
$\text { when } \mathrm{v}=0, \mathrm{a}=0$
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