# The acceleration of a object moving along the xaxis @ time t is given by a(t)=6t−2.If the velocity is 25 when t=3,the position is 10 when t=1,then find x(t). `a(t) = 6t-2`

The derivative of velocity function gives the acceleration function.

`(d(v(t)))/dt = a(t)`

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

By integrating both sides with respect to t

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

`v(t) = int a(t)dt`

`v(t) = int(6t-2)dt`

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

It is given...

`a(t) = 6t-2`

The derivative of velocity function gives the acceleration function.

`(d(v(t)))/dt = a(t)`

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

By integrating both sides with respect to t

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

`v(t) = int a(t)dt`

`v(t) = int(6t-2)dt`

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

It is given that when t = 3 then v(t) = 25

`25 = 3*3^2-2*3+C`

`25 = 21+C`

`C = 4`

`v(t) = 3t^2-2t+4`

The derivative of position function gives the velocity function.

`(d(x(t)))/dt = v(t)`

`dx(t) = v(t)dt`

By integrating both sides with respect to t.

`intdx(t) = intv(t)dt`

`x(t) = int v(t)dt`

`x(t) = int(3t^2-2t+4)dt`

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

It is given that then t = 1 then the position is 10.

`10 = 1-1+4+C1`

`C1 = 6`

So the position function is,

`x(t) = t^3-t^2+4t+6`

Approved by eNotes Editorial Team