The position of a particle travelling along x

Question -

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

1) the particle does not come to rest at all.

2) the particle comes to rest firstly at (3 - $\sqrt{7}$ ) s and then at (3 + $\sqrt{7}$ ) s

3) the speed of the particle at t= 2s is 18 m $s^{- 1}$

4) the acceleration of the particle at t= 2s is 6 m $s^{- 2}$

Options:

A . 2,3, 4 are correct

B . 2,3 are correct

C . All of them are correct

D . 1 and 2 are correct

Answer: Option A
:
A

$x_{t}$ = $t^{3}$ - 9 $t^{2}$ + 6t

v= $\frac{d x_{t}}{d t}$ = (3 $t^{2}$ - 18t + 6) m $s^{- 1}$

So, the particle comes to rest since this equation has two valid roots.

$\Rightarrow$ $t_{1}$ = (3 - $\sqrt{7}$ ) s and $\Rightarrow$ $t_{2}$ = (3 - $\sqrt{7}$ ) s

Substituting the value t = 2s in the equation for velocity, we get, v= = - 18 m $s^{- 1}$

The acceleration of the particle is a=6t - 18 m $s^{- 2}$

Substituting the value t = 2s in the equation for acceleration, we get, a= = - 6 m $s^{- 2}$

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