A particle starts from the origin and moves along the $X$-axis such that the velocity at any instant is given by $4t^{3} - 2t$, where $t$ is in second and velocity is in $ms^{- 1}$. What is the acceleration of the particle when it is 2 m from the origin?

Question

Dr. Abubacker · Accepted Answer

(c) $\frac{dx}{dt} = 4t^{3} - 2t\ $
or $dx = 4t^{3}dt - 2t\ dt$
Integrating, $x = \frac{4t^{4}}{4} - \frac{2t^{2}}{2} = t^{3} - t^{2}$
When $x = 2,t^{4} - t^{2} - 2 = 0,$
$t^{2} = \frac{- ( - 1) \pm \sqrt{1 + 8}}{2}$
or $t^{2} = \frac{1 \pm 3}{2} = 2\ $(ignoring – ve sign)
Again, $\frac{d^{2}x}{dt^{2}} = 12t^{2} - 2$
When $t^{2} = 2,$ acceleration=12$	imes 2 - 2 = 22ms^{- 2}$

Q. A particle starts from the origin and moves along the $X$-axis such that the velocity at any instant is given by $4t^{3} - 2t$, where $t$ is in second and velocity is in $ms^{- 1}$. What is the acceleration of the particle when it is 2 m from the origin?

Explanation

Reviewed by Dr. Abubacker

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