Deriving the Formula for Final Velocity in One Dimension

[SaltyBriefs]

Homework Statement

I'm trying to derive this formula, but I get stuck after I factor the t out.

V_{f}^{2} = V_{0}^{2} + 2a (y-y_{0})

Homework Equations

V_{f}^{2} = V_{0}^{2} + 2a (y-y_{0})

The Attempt at a Solution

1) y_{f} - y_{0} = (\frac{V_{0}+V_{f}}{2})t

2) y_{f} - y_{0} (\frac{1}{t})= (\frac{V_{0}+V_{f}}{2})t (\frac{1}{t})

3) V_{f}= (\frac{V_{0}+V_{f}}{2})t (\frac{1}{t})

4) ?

5) V_{y} = V_{0y}^{2} + 2a (y_{f}-y_{0})

 

[PeterO]

Homework Statement

I'm trying to derive this formula, but I get stuck after I factor the t out.

V_{f}^{2} = V_{0}^{2} + 2a (y-y_{0})

Homework Equations

V_{f}^{2} = V_{0}^{2} + 2a (y-y_{0})

The Attempt at a Solution

1) y_{f} - y_{0} = (\frac{V_{0}+V_{f}}{2})t

2) y_{f} - y_{0} (\frac{1}{t})= (\frac{V_{0}+V_{f}}{2})t (\frac{1}{t})

3) V_{f}= (\frac{V_{0}+V_{f}}{2})t (\frac{1}{t})

4) ?

5) V_{y} = V_{0y}^{2} + 2a (y_{f}-y_{0})

Firstly, I can only show this using symbols
v for your V_f - final velocity
u for your V_o - initial velocity
s for our y - y_o - displacement.
{partially from familiarity, and partly because it is easier to type}

So I am aiming at v² = u² + 2as

you are hopefully familiar with a couple of other motion equations

v = u +at & s = t*(v + u)/2 [you listed tis second one in line 1 of your solution]

These two are combined.

The first can be transposed to give

t = (v - u)/a

substitute for t in the second

s = (v-u)(v+u)/2a

so

(v-u)(v+u) = 2as

v² - u² = 2as

or

v² = u² + 2as

 

[WatermelonPig]

Alternatively:

a = dv/dt

a(ds) = ds/dt(dv)

int(ads) = int(vdv)

Assuming a = constant.

as = 1/2(v^2-u^2)

v^2 = u^2 + 2as