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).
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`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`
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