高専の数学解説

第7章 7.14

y=1002−(4t)2y = \sqrt{100^2 - (4t)^2}

y′=−32t21002−(4t)2=−16t1002−(4t)2y' = \frac{-32t}{2\sqrt{100^2-(4t)^2}} = -\frac{16t}{\sqrt{100^2-(4t)^2}}

OA = 60cm なので t = 15 秒のとき

y′=−480210000−3600=−2406400=−24080=−3y' = \frac{-480}{2\sqrt{10000-3600}} = \frac{-240}{\sqrt{6400}} = -\frac{240}{80} = -3 －3 cm/s

加速度は

y′′=−161002−(4t)2−16t⋅(−16t1002−(4t)2)1002−(4t)2=−16⋅(1002−(4t)2)+256t21002−(4t)21002−(4t)2y'' = -\frac{16\sqrt{100^2-(4t)^2} - 16t\cdot\left(-\dfrac{16t}{\sqrt{100^2-(4t)^2}}\right)}{100^2-(4t)^2} = -\frac{\dfrac{16\cdot(100^2-(4t)^2)+256t^2}{\sqrt{100^2-(4t)^2}}}{100^2-(4t)^2}

t = 15 のとき

y′′=−16⋅6400+256⋅225806400=−160000512000=−0.3125y'' = -\frac{\dfrac{16\cdot 6400 + 256\cdot 225}{80}}{6400} = -\frac{160000}{512000} = -0.3125 －0.3125 cm/s²

解説: r31bn1z