Ta có:
\(\Delta s=2v_0+\frac{1}{2}a.2^2-\left(4v_0+\frac{1}{2}a.4^2-2v_0-\frac{1}{2}a.2^2\right)\\ \Rightarrow-4a=4\\ \Rightarrow a=-1\left(m\text{/}s^2\right)\)
Suy ra \(v_0=10\left(m\text{/}s\right)\Rightarrow s=v_0t+\frac{1}{2}at^2=10.10+\frac{1}{2}.\left(-1\right).10^2=50\left(m\right)\)