움직인 거리

시각 $t \left( t \geq 0 \right)$에서의 점 $\mathrm{P}$의 위치를 $x \left( t \right)$라 하자. $x \left( t \right)$$\displaystyle = x \left( 0 \right) + \int _{0} ^{t} {} v \left( t \right) dt$ $\displaystyle = 16 + \int _{0} ^{t} {} 3 t \left( a - t \right) dt$ $\displaystyle = 16 + \int _{0} ^{t} {} \left( - 3 t ^{2} + 3 at \right) dt$ $\displaystyle = 16 + \left[ \begin{array}{l} \begin{matrix} \\ \end{matrix} - t ^{3} + \frac{3}{2} at ^{2} \end{array} \right] _{0} ^{t}$ $\displaystyle = - t ^{3} + \frac{3}{2} at ^{2} + 16$ 시각 $t = 2 a$에서 점 $\mathrm{P}$의 위치가 $\mathrm{0}$이므로 $x \left( 2 a \right)$$\displaystyle = - \left( 2 a \right) ^{3} + \frac{3}{2} a \times \left( 2 a \right) ^{2} + 16$ $= 16 - 2 a ^{3} = 0$ $a ^{3} = 8$, $a = 2$ $v \left( t \right) = 3 t \left( 2 - t \right) = - 3 t ^{2} + 6 t$ 따라서 시각 $t = 0$에서 $t = 5$까지 점 $\mathrm{P}$가 움직인 거리는 $\displaystyle \int _{0} ^{5} {} \left| v \left( t \right) \right| dt$ $\displaystyle = \int _{0} ^{5} {} \left| - 3 t ^{2} + 6 t \right| dt$ $\displaystyle = \int _{0} ^{2} {} \left( - 3 t ^{2} + 6 t \right) dt + \int _{2} ^{5} {} \left( 3 t ^{2} - 6 t \right) dt$ $\displaystyle = \left[ \begin{array}{l} \begin{matrix} \\ \end{matrix} \begin{matrix} \\ \end{matrix} - t ^{3} + 3 t ^{2} \end{array} \right] _{0} ^{2} + \left[ \begin{array}{l} \begin{matrix} \\ \end{matrix} \begin{matrix} \\ \end{matrix} t ^{3} - 3 t ^{2} \end{array} \right] _{2} ^{5}$ $= \left\{ \left( - 8 + 12 \right) - 0 \right\} + \left\{ \left( 125 - 75 \right) - \left( 8 - 12 \right) \right\}$ $= 58$