# Derive the following kinematics equation using the position and velocity equations

1. Dec 25, 2012

### tahayassen

1. The problem statement, all variables and given/known data

$${ y }_{ f }={ y }_{ i }+{ v }_{ yi }t+\frac { 1 }{ 2 } { a }_{ y }{ t }^{ 2 }\\ { v }_{ yf }={ v }_{ yi }+{ a }_{ y }t$$

Derive

$${ { v }_{ yf } }^{ 2 }={ { v }_{ yi } }^{ 2 }+2{ a }_{ y }({ y }_{ f }-{ y }_{ i })$$

2. The attempt at a solution

This isn't an actual homework question; I'm just cross-posting this thread because I'm terribly impatient and there is much more traffic here.

2. Dec 25, 2012

### Staff: Mentor

Re: Derive the following kinematics equation using the position and velocity equation

Solve for t in the second equation, and substitute it into the first equation.

3. Dec 25, 2012

### tahayassen

Re: Derive the following kinematics equation using the position and velocity equation

Thanks. That derivation took much longer than expected.

Last edited: Dec 25, 2012
4. Dec 25, 2012

### ehild

Re: Derive the following kinematics equation using the position and velocity equation

You can connect the equation to work-energy theorem. The work done on a particle is equal to the change of its kinetic energy. Work done is displacement times force. Force is F=ma. $$W=(y_f-y_i)ma=1/2(mv_f^2-mv_i^2)$$. Cancel m.

But you can derive the formula easily if you remember that the displacement is average velocity multiplied by time. $$Δy=\frac{1}{2}(v_i+v_f)t$$. Substitute $t=\frac{v_f-v_i}{a}$ for t.
$$y_f-y_i=\frac{1}{2}(v_i+v_f)\frac{v_f-v_i}{a}=\frac{v_f^2-v_i^2}{2a}$$

ehild