RRB Paramedical Staff Nurse Model Paper 3

Question : 92 of 100

Marks: +1, -0

The position of a particle travelling along x - axis is given by

xt = t^3 - 9 t^2 + 6t

where xt 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 - √7)

s and then at

(3 + √7)

s
3) the speed of the particle at

t= 2s

18 ms^{−1}

4) the acceleration of the particle at

t= 2s

6ms^{−2}

Solution:

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

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

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

⇒ t_{1}=(3-√{7}) {s}

and

⇒ t_{2}=(3-√{7}) {s}

Substituting the value

t=2 s

in the equation for velocity, we get,

v=-18 {m}

s^{-1}

The acceleration of the particle is a

=6 t-18 {ms}^{-2}

Substituting the value

t=2 s

in the equation for acceleration, we get,

a=6{ms}^{-2}

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