The position of a particle moving along x-axis given by x=(-2 t3+3 R2+5) m. The acceleration of particle at the instant its velocity becomes zero is

Question

The position of a particle moving along x-axis given by x=(-2 t3+3 R2+5) 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

Tardigrade Admin · Accepted Answer

x=(-2 t3+3 t2+5) m ⇒ (d x/d t)=-6 t2+6 t=v ⇒ (d2 x/d t2)=-12 t+6 (for .v=0,6 t=6 t2 ⇒ t=1 s ) .a|t=1 s=-12+6=-6 ms -2

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

$12 m / s^{2}$

$- 12 m / s^{2}$

$- 6 m / s^{2}$

Zero

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

Q. The position of a particle moving along x-axis given by x=(−2t3+3R2+5)m. The acceleration of particle at the instant its velocity becomes zero is

12m/s2

−12m/s2

−6m/s2