A particle initially at rest is moving along | vedclass

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(C) $1$. For the first $3 \,s$, the particle starts from rest $(u=0)$ with acceleration $a_1 = 2 \,m/s^2$.
$2$. Velocity at $t=3 \,s$ is $v = u + a_1

Vedclass · Accepted Answer

$3(1+\sqrt{2}) \,s$

Vedclass · Answer

(C) $1$. For the first $3 \,s$, the particle starts from rest $(u=0)$ with acceleration $a_1 = 2 \,m/s^2$.
$2$. Velocity at $t=3 \,s$ is $v = u + a_1 t = 0 + 2(3) = 6 \,m/s$.
$3$. Displacement at $t=3 \,s$ is $s_1 = ut + 0.5 a_1 t^2 = 0 + 0.5(2)(3^2) = 9 \,m$.
$4$. After $t=3 \,s$, the acceleration reverses, so $a_2 = -2 \,m/s^2$. Let the additional time be $t'$.
$5$. The position at time $t'$ is $s = s_1 + vt' + 0.5 a_2 (t')^2$.
$6$. For the particle to return to its initial position, $s=0$.
$7$. $0 = 9 + 6t' - 0.5(2)(t')^2 \Rightarrow 0 = 9 + 6t' - (t')^2 \Rightarrow (t')^2 - 6t' - 9 = 0$.
$8$. Using the quadratic formula $t' = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{6 \pm \sqrt{36 - 4(1)(-9)}}{2} = \frac{6 \pm \sqrt{72}}{2} = 3 \pm 3\sqrt{2}$.
$9$. Since $t' > 0$, $t' = 3 + 3\sqrt{2} = 3(1+\sqrt{2}) \,s$.
$10$. Total time $T = 3 + t' = 3 + 3 + 3\sqrt{2} = 6 + 3\sqrt{2} = 3(2+\sqrt{2}) \,s$.

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