PLease explain an equation

definition of acceleration:
a = (v-u) / t ..........................................(1)

definition of velocity:
V = s/ t ...............................................(2)

definition of average velocity:
V = (v+u) / 2 ........................................(3)
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Combine (2) & (3),

s / t = (v+u) / 2
s = t [(v+u) / 2]
s = (v +u) t / 2 .......................................(4)
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From (1),

a = (v-u) / t
at = v -u
v = at +u ................................................(5)
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Put (5) into (4),

s = (v +u)t / 2
s = [(at +u) +u] (t / 2)
s = (at +2u) (t / 2)
s = [at (t/2)] + [2u (t/2)]
s = 1/2 at^2 + ut
s = ut + 1/2 at^2 ....................................(result)

We should explain it by means of integration.

Firstly, by definition,

Velocity, v = ∫a dt, where a is acceleration, t is time

It results in: v = at + c, where c is a constant

2008-08-24 14:03:20 補充：
Taking at t = 0, v = u. So, c = u

Therefore, v = u + at

Now, displacement, s = ∫v dt

s = ∫(u + at)dt

s = ut + 1/2 at2 + d, where d is a constant

2008-08-24 14:03:37 補充：
When t = 0, s = 0. So, d = 0

Therefore, we get s = ut + 1/2 at2


Note that this equation is only true when acceleration, a is uniform.

If acceleration is varying, you should use integration to work out the relations among s, v, t and a.

Q: why at^2 has to be divided by 2

The displacement s is the area under a velocity-time graph. For uniform acceleration, such a graph is a straight line with slope equla to the acceleration a.

THus, the area under the straight line is the area of a trapezium, with lengths of the top-side = u, and the bottom-side = v, and height t.

The trapezium can be separated into a rectangle of area = u.t and a triangle with area given by (1/2)x base x height = (1/2).t.(at) = (1/2)a.t^2

Therefore, the (1/2) in fact comes from the area of the triangle that constitutes part of the total area of the trapezium, which represents s