Question

how we derived vf=vi+at

Answer 1

Acceleration by definition is
\[a=\frac{ \Delta V }{ \Delta t }\]
\[\Delta V= V_f-V_i\]
so
\[a=\frac{ V_f-V_i }{ \Delta t }\]
multiply and solve for Vf ;)

Answer 2

another approach is by graphing V and t, the relationship is linear, acceleration is constant...the slope of the line is acceleration\[a=\frac{\Delta V}{\Delta t}=\frac{v_2-v_1}{t_2-t_1}\]if we let \(v_2=v_f\) and \(v_1=v_i\) and \(t_2-t_1=1\) then\[a=\frac{v_f-v_i}{t}\]\[v_f=v_i+at\]

Answer 3

typo error: it should be \(t_2-t_1=t\)