Question

Derive equation of motion s=ut+1/2at²​

Answer 1

Recall the definitions of

• average velocity:

v[ave] = ∆x/∆t = (x[final] - x[initial])/t

Take the initial position to be the origin, so x[initial] = 0, and we simply write x[final] = s. So

v[ave] = s/t

• average acceleration:

a[ave] = ∆v/∆t = (v[final] - v[initial])/t

Assume acceleration is constant (a[ave] = a). Let v[initial] = u and v[final] = v, so that

a = (v - u)/t

Under constant acceleration, the average velocity is also given by

v[ave] = (v[final] + v[initial])/2 = (v + u)/2

Then

v[ave] = s/t = (v + u)/2   ⇒   s = (v + u) t/2

and

a = (v - u)/t   ⇒   v = u + at

so that

s = ((u + at) + u) t/2

s = (2u + at) t/2

s = ut + 1/2 at²