2-2 할리데이 11판 솔루션 일반물리학

$$ \begin{cases} v_0 &= 3.70\times10^5\ut{m/s}\\ L &= 2.30\ut{cm}\\ v &= 4.40\times10^6\ut{m/s} \end{cases} $$ $$ \begin{aligned} 2aS&=v^2-{v_0}^2,\\ \end{aligned} $$ $$ \begin{aligned} a&=\frac{(4.40\times10^6\ut{m/s})^2-{(3.70\times10^5\ut{m/s})}^2}{2(2.30\ut{cm})}\cdot\frac{100\ut{cm}}{1\ut{m}} \\ &=\frac{9611550000000000}{23}\ut{m/s^2}\\ &\approx4.1789347826086956\times 10^{14}\ut{m/s^2}\\ &\approx4.18\times 10^{14}\ut{m/s^2}\\ \end{aligned} $$