MCQ
Relation between velocity and displacement is $v = x^2$. Find acceleration at $x = 3m$ :- ............. $\mathrm{m/s}^{2}$
- A$6$
- B$27$
- ✓$54$
- D$0$
$\frac{\mathrm{dV}}{\mathrm{dx}}=2 \mathrm{x}$
$\frac{\mathrm{vdv}}{\mathrm{dx}}=\mathrm{v} \cdot 2 \mathrm{x}$
$a=2 v x=2 x^{2} x$
${a=2 x^{3}}$
${a=2(3)^{3}}$
${a=54 \mathrm{m} / \mathrm{s}^{2}}$
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