Why is the 1/2 term necessary in physics equations?

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The "1/2" term in physics equations, such as d = 1/2at² and E = 1/2mv², arises from the need to account for average values during uniformly accelerated motion. In the equation d = vt, "v" represents average velocity, while "v" in v = at denotes instantaneous velocity. The average velocity during constant acceleration from rest is v' = (v₀ + v)/2, leading to the inclusion of the "1/2" factor in the displacement equation. This factor is essential for accurately calculating displacement and kinetic energy when dealing with changing velocities.

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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?
 
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DecayProduct said:
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?

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

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