For a constant acceleration environment we want to find equations to link:
Initial speed: \(v_{t_0}\)
End speed: \(v_{t_1}\)
Time: \(t_1-t_0\)
Acceleration: \(a\)
Displacement \(s_{t_1}-s_{t_0}\)
These are the following, and are derived below.
\(v_{t_1}=a(t_1-t_0)+v_{t_0}\)
\((s_{t_1}-s_{t_0})=v_{t_0}(t_1-t_0)+\dfrac{1}{2}a(t_1-t_0)^2\)
\((s_{t_1}-s_{t_0})=v_{t_1}(t_1-t_0)-\dfrac{1}{2}a(t_1-t_0)^2\)
\(v_{t_1}^2= v_{t_0}^2+2a(s_{t_1}-s_{t_0})\)
\((s_{t_1}-s_{t_0})=(t_1-t_0)\dfrac{v_{t_1}+v_{t_1}}{2}\)
This equation is:
\(v_{t_1}=a(t_1-t_0)+v_{t_0}\)
To derive this start with:
\(v_t:=\dfrac{\delta s_t}{\delta t}\)
\(a:=\dfrac{\delta v_t}{\delta t}\)
If acceleration is constant, then
\(\dfrac{\delta v_t}{\delta t}=a\)
\(v_t=\int a dt +v_0\)
\(v_t=at+v_0\)
This equation is:
\((s_{t_1}-s_{t_0})=v_{t_0}(t_1-t_0)+\dfrac{1}{2}a(t_1-t_0)^2\)
To derive this start with:
\(v:=\dfrac{\delta s_t}{\delta t}\)
Then:
\(\dfrac{\delta s_t}{\delta t}=at+v_0\)
\(s_t=\dfrac{1}{2}at^2+v_0t+s_0\)
\((s_t-s_0)=v_0t+\dfrac{1}{2}at^2\)
This equation is:
\((s_{t_1}-s_{t_0})=v_{t_1}(t_1-t_0)-\dfrac{1}{2}a(t_1-t_0)^2\)
To derive this start with:
\(v_t=at+v_0\)
\((s_t-s_0)=t\dfrac{v_t+v_0}{2}\)
So:
\(v_0=v_t-at\)
\(v_0=\dfrac{2}{t}(s_t-s_0)- v_t\)
\(v_t-at=\dfrac{2}{t}(s_t-s_0)- v_t\)
\((s_t-s_0)=v_tt-\dfrac{1}{2}at^2\)
This equation is:
\(v_{t_1}^2= v_{t_0}^2+2a(s_{t_1}-s_{t_0})\)
To derive this start with:
\(v_t=at+v_0\)
\((s_t-s_0)=t\dfrac{v_t+v_0}{2}\)
So:
\(t=\dfrac{v_t-v_0}{a}\)
\(t=2\dfrac{s_t-s_0}{v_t+v_0}\)
\(\dfrac{v_t-v_0}{a}=2\dfrac{s_t-s_0}{v_t+v_0}\)
\((v_t-v_0)(v_t+v_0)=2a(s_t-s_0)\)
\(v^2_t= v^2_0+2a(s_t-s_0)\)
This equation is:
\((s_{t_1}-s_{t_0})=(t_1-t_0)\dfrac{v_{t_1}+v_{t_1}}{2}\)
To derive this start with:
\(v_t=at+v_0\)
\(s_t-s_0=\dfrac{1}{2}at^2+v_0t\)
So:
\(a=\dfrac{v_t-v_0}{t}\)
\(a=\dfrac{2[(s_t-s_0)-v_0t]}{t^2}\)
\(\dfrac{v_t-v_0}{t}=\dfrac{2[(s_t-s_0)-v_0t]}{t^2}\)
\(t(v_t-v_0)=2[(s_t-s_0)-v_0t]\)
\(t(v_t+v_0)=2(s_t-s_0)\)
\((s_t-s_0)=t\dfrac{v_t+v_0}{2}\)