# A particle is moving with the given data. Find the position of the particle. A particle is moving with the given data. Find the position of the particle. a(t) = t2 − 8t + 4,  s(0) = 0,  s(1) = 20   s(t)=_____________? `a(t) = t^2-8t+4`

The derivative of velocity function v(t) is the acceleration function.

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

`dv(t) = a(t)dt`

`intdv(t) = inta(t)dt`

`v(t) = inta(t)dt`

`v(t) = int(t^2-8t+4)dt`

`v(t) = t^3/3-4t^2+4t+C1` where C1 is a constant.

The derivative of position function is the velocity function.

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

...

`a(t) = t^2-8t+4`

The derivative of velocity function v(t) is the acceleration function.

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

`dv(t) = a(t)dt`

`intdv(t) = inta(t)dt`

`v(t) = inta(t)dt`

`v(t) = int(t^2-8t+4)dt`

`v(t) = t^3/3-4t^2+4t+C1` where C1 is a constant.

The derivative of position function is the velocity function.

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

`ds(t) = v(t)dt`

`intds(t) = intv(t)dt`

`s(t) = intv(t)dt`

`s(t) = int(t^3/3-4t^2+4t+C1)dt`

`s(t) = t^4/12-4t^3/3+2t^2+C1t+C2` where C2 is a constant.

`s(t) = t^4/12-4t^3/3+2t^2+C1t+C2`

It is given that;

`s(0) = 0`

`s(1) = 20 `

`0 = 0^4/12-4*0^3/3+2*0^2+C1*0+C2`

`0 = C2`

`20 = 1^4/12-4*1^3/3+2*1^2+C1*1+0`

`20 = 3/4+C1`

`C1 = 77/4`

So the position function is;

`s(t) = t^4/12-(4t^3)/3+2t^2+(77t)/4`

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