Answer
$$v(0)=1.$$
Work Step by Step
Since $r(t)=\lt 0,e^t, -\cos(2t)\gt$, then the velocity vector is given by
$$r'(t)=\lt 0,e^t, 2\sin(2t)\gt$$
and the acceleration vector is given by
$$r''(t)=\lt 0,e^t, 4\cos(2t)\gt.$$
Moreover, the speed at $t=0$ is
$$v(0)=\|r'(0)\|=\sqrt{0+1+0}=1.$$
