anonymous one year ago Mhm

1. anonymous

$\text{Total area} = X + Y \implies \frac{ 1 }{ 2 }t(\vec v_f- \vec v_i) +\vec v_i t$

2. anonymous

Where total area is displacement

3. anonymous

$\vec d = \frac{ \vec v_f t }{ 2 }- \frac{ \vec v_i t }{ 2 }+v_i t$ $\vec d = \frac{ \vec v_f t }{ 2 }+\frac{ \vec v_i t }{ 2 } = \left( \frac{ \vec v_f + \vec v_i }{ 2 } \right)t$ $\text{Therefore} ~ \vec d = \left( \frac{ \vec v_f + \vec v_i }{ 2 } \right)t$

4. anonymous

Ah, so now similarly can derive the other equations of uniform motion, using just the graphs :).

5. anonymous

|dw:1437888152528:dw|Using above being derived from

6. anonymous

Well we can derive $\vec d = \vec v_i t + \frac{ 1 }{ 2 } \vec at^2$ and then to find $\vec v_f ^2= \vec v_i^2+ 2a \vec d$ we can substitute $\vec a = \frac{ \vec v_f - \vec v_i }{ t }$ into $\vec d = \vec v_i t + \frac{ 1 }{ 2 } \vec a t^2$