# The position of a particle moving along a straight line at any time t is given by s(t)=(2t^3)-(4t^2)+2t-1. What is the acceleration of the particle when t=2?

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The position vector is given as,

`s(t) = 2t^3-4t^2+2t-1`

This gives the displacement of the particle at a given time t.

We know,

Velocity= `v = (ds)/(dt)`

Therefore by differentiating wrt t,

`v(t) = 6t^2-8t+2`

Also we know that,

Acceleration = `a = (dv)/(dt)`

Therefore by differentiating wrt t,

`a(t) = 12t-8`

Therefore the acceleration of the particle at time t is given by,

`a(t) = 12t -8`

The acceleration at t=2,

`a(2) = 12 xx 2 -8`

`a(2) = 24-8 = 16`

**Therefore the acceleration at t =2 is 16.**