Question

The position of a particle travelling along x axis is given by $x_1=t^3-9t^2+6t$ where $x_t$ is in $cm$ and $t$ is in second. Then

• A. the body comes to rest firstly at $(3-\sqrt{7})$ and then at $(3+\sqrt{7})$
• 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 velocity to the second zero of velocity is $-74cm$.
• D. the particle reverses its velocity at $(3-\sqrt{7})$ s and then at $(3+\sqrt{7})$ s and has a negative velocity for $(3-\sqrt{7})$ < t < $(3+\sqrt{7})$

Answer 1

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

The correct options are
A the body comes to rest firstly at $(3-\sqrt{7})$ and then at $(3+\sqrt{7})$
C the total displacement of the particle in travelling from the first zero of velocity to the second zero of velocity is $-74cm$.
D the particle reverses its velocity at $(3-\sqrt{7})$ s and then at $(3+\sqrt{7})$ s and has a negative velocity for $(3-\sqrt{7})$ < t < $(3+\sqrt{7})$

$x_t=t^3-9t^2+6t$
$V=\frac{d_{X_t}}{dt}=(3t^2-18t+6)cm{s}^{-1}$
For body to be at rest $v=0$
$\Rightarrow t=(3\pm \sqrt{7})s$
$\Rightarrow t_1=(3-\sqrt{7})s$ and $\Rightarrow t_2=(3+\sqrt{7})s$
Displacement of particle between $t_1$ and $t_2$ is ${x_t}_2-{x_t}_1$
$({(3+\sqrt{7})}^3-9{(3+\sqrt{7})}^2+6(3+\sqrt{7}))-({(3-\sqrt{7})}^3-9{(3-\sqrt{7})}^2+6(3-\sqrt{7}))=-74$
$v<0for(3-\sqrt{7})<t<(3+\sqrt{7})$