The velocity of the particle is given by
[tex]v(t)=v(0)+\displaystyle\int_0^ta(u)\,\mathrm du[/tex]
Since [tex]a(t)=-4\sin2t[/tex] and [tex]v(0)=7[/tex], we get
[tex]\displaystyle\int_0^ta(u)\,\mathrm du=\int_0^t-4\sin2u\,\mathrm du=2\cos2u\bigg|_0^t=2\cos2t-2[/tex]
[tex]\implies v(t)=2\cos2t+5[/tex]
Similarly, the position function is obtained via
[tex]x(t)=x(0)+\displaystyle\int_0^tv(u)\,\mathrm du[/tex]
We know [tex]v(t)[/tex] and we're told that [tex]x(0)=0[/tex], so
[tex]\displaystyle\int_0^tv(u)\,\mathrm du=\int_0^t(2\cos2u+5)\,\mathrm du=\sin2u+5u\bigg|_0^t=\sin2t+5t[/tex]
[tex]\implies x(t)=\sin2t+5t[/tex]
making the answer A.
