Đáp án:
\({v^2} - v_0^2 = 2as\)
Giải thích các bước giải:
Ta có:
\(v = {v_0} + at \Rightarrow t = \dfrac{{v - {v_0}}}{a}\)
Thay vào:
\(\begin{array}{l}
s = {v_0}t + \dfrac{1}{2}a{t^2} = {v_0}.\dfrac{{v - {v_0}}}{a} + \dfrac{1}{2}a.{(\dfrac{{v - {v_0}}}{a})^2}\\
\Rightarrow s = \dfrac{{{v_0}v - {v^2}}}{a} + \dfrac{1}{2}.\dfrac{{{v^2} - 2v.{v_0} + v_0^2}}{a}\\
\Rightarrow s = \dfrac{{2{v_0}v - 2{v_0}^2}}{{2a}} + \dfrac{{{v^2} - 2v.{v_0} + v_0^2}}{{2a}} = \dfrac{{{v^2} - v_0^2}}{{2a}}\\
\Rightarrow {v^2} - v_0^2 = 2as(dpcm)
\end{array}\)
