Question

The position of particle travelling along x-axis is given by $x_t=t^3–9t^2+6t$ where $x_t$ is in $\text{cm}$ and $t$ is in second. Then–

• A. The body comes to rest first at $(3–\sqrt{7})s$ and then at $(3+\sqrt{7})s$
• B. The total displacement of the particle in travelling from the first zero of velocity to the second zero of velocity is zero
• C. The total displacement of the particle in travelling from the first zero of the velocity to the second zero of velocity is $–74cm$
• D. The particle reverses it’s velocity at $(3–\sqrt{7})s$ and then at $(3+\sqrt{7})s$ and has a negative velocity for $(3–\sqrt{7})s<t<(3+\sqrt{7})s$

Answer 1

Answer: (A), (C), (D)

The correct options are
A The body comes to rest first at $(3–\sqrt{7})s$ and then at $(3+\sqrt{7})s$
C The total displacement of the particle in travelling from the first zero of the velocity to the second zero of velocity is $–74cm$
D The particle reverses it’s velocity at $(3–\sqrt{7})s$ and then at $(3+\sqrt{7})s$ and has a negative velocity for $(3–\sqrt{7})s<t<(3+\sqrt{7})s$

$x_t=t^3-9t^2+6t$
$v=\frac{d{x}_t}{dt}=(3t^2+18t+6){\text{cms}}^{-1}$
for $v=0$ $\Rightarrow t=(3\pm \sqrt{7})s$
$\Rightarrow t_1=(3-\sqrt{7})s$ and $\Rightarrow t_2=(3+\sqrt{7})s$
Displacement of particle between $t_1$ and $t_2$ is
$x_{t_2}-x_{t_1}$
$=\left(\right(3+\sqrt{7}{)}^3-9(3+\sqrt{7}{)}^2+6(3+\sqrt{7}\left)\right)-\left(\right(3-\sqrt{7}{)}^3-9(3-\sqrt{7}{)}^2+6(3-\sqrt{7}\left)\right)=-74$
$v<0$ for $(3-\sqrt{7})<t<(3+\sqrt{7})$