# Linear motion equations

1. Mar 10, 2012

### jshpark

how were these equations derived or found?

x = xo + vot + 1/2at2

I tried to use v=(xf-xi)/(tf-ti) and a=(vf-vi)/(tf-ti)

but I don't know where the 1/2 comes from

also this one:
v2 = vo2 + 2a(xf-xi)

2. Mar 11, 2012

### Fredrik

Staff Emeritus
The first one is what you get if you integrate the equation that describes constant acceleration twice. If the acceleration is constant (=a), we have
$$x''(t)=a$$ This implies that there's a constant $v_0$ such that $$x'(t)=at+v_0.$$ The constant is denoted by $v_0$ because $x'(t)$ is the velocity at time t, and the result we just found implies that $x'(0)=v_0$, so $v_0$ is the velocity at time 0. The result we found also implies that there's a constant $x_0$ such that
$$x(t)=\frac 1 2 at^2+v_0t+x_0.$$ The constant is denoted by $x_0$ because it's equal to $x(0)$, the position at time 0.

The other one follows from the result for the velocity above, and
$$v(t)-v_0=\int_0^t x''(t)dt=at.$$ These results imply that
$$v(t)^2=(v_0+at)^2=v_0{}^2+a^2t^2+2v_0at=v_0{}^2+2a\bigg(\frac 1 2 at^2+v_0t\bigg)=(v_0)^2+2a(x(t)-x_0).$$

I think the formula you wrote down only holds if $v(t_i)=v_0$, so you may want to check your book to see if it says exactly what you wrote, or something slightly different.

Last edited: Mar 11, 2012