uniformly-accelerated motion

クリスカ^{∗}

2010 年 12 月 21 日

Let x_{0} is initial displacement. x is displacement. v_{0} is initial velocity. v is final velocity. a is uniform acceleration. t is time.

So

a = \frac{dv}{dt} i.e. v = at + v_{0}

v = at + v_{0} ⟺ \frac{dx}{dt} = at + v_{0} i.e. x = \frac{1}{2}at^{2} + v_{0}t + x_{0}

v^{2} − v_{0}^{2} = (v − v_{0})(v + v_{0})

= at × ((at + v_{0}) + v_{0})

= 2a (\frac{1}{2}at^{2} + v_{0}t)

= 2a(x − x_{0})

Hence we get following

\{ \begin{matrix}v & = & at + v_{0} \\ x & = & \frac{1}{2}at^{2} + v_{0}t + x_{0} \\ v^{2} − v_{0}^{2} & = & 2a(x − x_{0}) \end{matrix}

∗ k-nakagawa@h6.dion.ne.jp