Yes. Let $$\tau^{\leq0}A^\bullet:= \cdots\to A^{-2}\to A^{-1}\to\ker(d^0)\to0\to\cdots$$ be the ususal truncation of $A^\bullet$. Then the mapping cone of the inclusion map $\tau^{\leq0}A^\bullet\to A^\bullet$ is homotopy equivalent to $$\rho^{>0}A^\bullet:=\cdots\to0\to\ker(d^0)\to A^0\to A^1\to\cdots,$$ so there is a distinguished triangle $$\tau^{\leq0}A^\bullet\to A^\bullet\to\rho^{>0}A^\bullet\to(\tau^{\leq0}A^\bullet)[1]$$ in $K(\mathcal{A})$.
Or another way to see this triangle: If the contractible complex $$K^\bullet:=\cdots\to0\to\ker(d^0)\xrightarrow{\sim}\ker(d^0)\to0\to\cdots$$ is concentrated in degrees $-1$ and $0$, then there is a degreewise split short exact sequence $$0\to\tau^{\leq0}\to A^\bullet\oplus K^\bullet\to\rho^{>0}\to0.$$