Luôn có: ${{\left( \frac{v}{{{v}_{\max }}} \right)}^{2}}+{{\left( \frac{a}{{{a}_{\max }}} \right)}^{2}}=1$ (*)
Khi $\left| v \right|=0,6{{v}_{\max }}$, (*) →${{\left( 0,6 \right)}^{2}}+{{\left( \frac{a}{{{a}_{\max }}} \right)}^{2}}=1\to \left| a \right|=0,8{{a}_{\max }}$.
