Question

The position of a particle moving along $x$-axis given by $x=\left(-2 t^{3}+3 R^{2}+5\right) m$. The acceleration of particle at the instant its velocity becomes zero is

• A. $12\, m / s ^{2}$
• B. $-12\, m / s ^{2}$
• C. $-6\, m / s ^{2}$
• D. Zero

Answer 1

Answer: (C)

$x=\left(-2 t^{3}+3 t^{2}+5\right) m$
$\Rightarrow \frac{d x}{d t}=-6 t^{2}+6 t=v$
$\Rightarrow \frac{d^{2} x}{d t^{2}}=-12 t+6$
(for $\left.v=0,6 t=6 t^{2} \Rightarrow t=1 s \right)$
$\left.a\right|_{t=1\, s}=-12+6=-6\, ms ^{-2}$