# Origin of terms in equations

1. Jan 16, 2010

### DecayProduct

I'm curious as to the origin of the "1/2" in some of the basic equations in physics. For example, d=1/2at$$^{2}$$. If d=vt, and v=at, then d=at$$^{2}$$, yet in reality we need the "1/2". Why?

2. Jan 16, 2010

### Staff: Mentor

In d = vt, "v" is the average velocity during the time interval between time 0 and time t.

In v = at, "v" is the instantaneous velocity at time t.

With constant acceleration starting from rest, the average velocity during the time interval between time 0 and time t is v/2 = at/2 (where "v" is again the final velocity at time t).

3. Jan 16, 2010

### Feldoh

In $$d = vt + d_0$$ your velocity is changing with time. The velocity in this equation should be the average velocity, call it v'.

$$v' = \frac{v_0+v}{2} => d = (v')t$$

In $$v = at + v_0$$ you're assuming constant acceleration so the average, a' is equal to a.

$$d = v't + d_0 = \frac{(v_0+v)t}{2} + d_0= \frac{(v_0 + at + v_0)t}{2} + d_0$$

So simplifying we get:

$$d = d_0 + v_0 t + \frac{1}{2}at^2$$

4. Jan 16, 2010

### DecayProduct

Thanks alot, that makes sense now, being the average. So how does it arise in E=1/2mv$$^{2}$$?

5. Jan 16, 2010

### Feldoh

Kinetic energy is for instantaneous velocity. Just thinking about it you should get a contradiction if you take an average energy vs. an average velocity due to the velocity being squared.