`a(t)=10sin(t)+3cos(t)`

`a(t)=v'(t)`

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

`v(t)=int(10sin(t)+3cos(t))dt`

`v(t)=-10cos(t)+3sin(t)+c_1`

`v(t)=s'(t)`

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

`s(t)=int(-10cos(t)+3sin(t)+c_1)dt`

`s(t)=-10sin(t)+3(-cos(t))+c_1t+c_2`

`s(t)=-10sin(t)-3cos(t)+c_1t+c_2`

Let's find constants c_1 and c_2 , given s(0)=0 ans s(2pi)=12

`s(0)=0=-10sin(0)-3cos(0)+c_1(0)+c_2`

`0=-3+c_2`

`c_2=3`

`s(2pi)=12=-10sin(2pi)-3cos(2pi)+c_1(2pi)+3`

`12=-10(0)-3(1)+2pic_1+3`

`12=2pic_1`

`c_1=6/pi`

`:.` position of the particle is given by `s(t)=-10sin(t)-3cos(t)+(6t)/pi+3`

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