Derive the relations:

(i) \(v^2-u^2=2as\)

(ii) \(v=u+at\)

(iii) \(s=ut+\frac{1}{2}at^2\)

(i) Let 

x₁, v₁ = position and velocity of the object at time t₁ respectively.

x₂, v₂ = position and velocity of the object at time t₂.

a = uniform acceleration of the object. 

v² - u² = 2as

Derivation – acceleration is given by

= \(\frac{v_2-v_1}{t_2-t_1}\),

where v₁ and v₂, t₁ and t₂ are as in …(i)

or t₂ − t₁ = \(\frac{v_2-v_1}{a}\) …(i)

Since, x₂ − x₁ = v₁ (t₂ − t₁ ) + \(\frac{1}{2}\) a(t₂ − t₁)² …(ii)

From (i) and (ii), we get

\(x_2-x_1=v_1\frac{v_2-v_1}{a}+\frac{1}{2}a{[\frac{v_2-v_1}{a}]}^2\)

= \(\frac{v_1v_2-v^2_1}{a}+\frac{v_2^2v_1^2-2v_1v_2}{2a}\)

= \(\frac{2v_1v_2-2v^2_1+v_1^2+v^2_2-2v_1v_2}{2a}\)

\(x_2-x_1=\frac{v_2^2-v_1^2}{2a}\)   ...(iii)

Or  \(v_2^2-v_1^2=2a(x_2-x_1)\)  ...(iv)

Now if v₁ = u at t₁ = 0

v₂ = v at t₂ = t   ...(v)

x₂ - x₁ = s

Then from (iv) and (v), we get

v^₂ - u^₂ = 2as  ...(vi)

(ii) v =u + at

Derivation – By definition of acceleration,

a = \(\frac{v_2-v_1}{t_2-t_1}\)

or  v₂ - v₁ =a( t₂ - t₁ )

or  v_{2 =} v₁ + a( t₂ - t₁ )  ...(i)

where v₁ and v₂ are the velocities of an object at times t₁ and t₂ respectively.

If v₁ = u (initial velocity of the object)

at t₁ = 0

v₂ = v(final velocity of the object)

at t₂ = t

Then (i) reduce to v = u + at

(iii) \(s=ut +\frac{1}{2}at^2\)

Derivation – Also let v_av = average velocity in t₂ − t₁ interval.

By definition

\(v_{av}=\frac{x_2-x_1}{t_2-t_1}\)

or  \(x_2-x_1=v_{av}(t_2-t_1)\)   ...(i)

Since, v_av = \(\frac{v_1+v_2}{2}\)  ...(ii)

∴ From eqns. (i) and (ii),

\(x_2-x_1=\frac{v_1+v_2}{2}(t_2-t_1)\)   ...(iii)

Also, we know that

\(v_2=v_1+a(t_2-t_1)\)   ...(iv)

∴ From eqns. (iii) and (iv),

\(x_2-x_1=\frac{1}{2}[v_1+v_2+a(t_2-t_1)](t_2-t_1)\)

= \(v_1(t_2-t_1)+\frac{1}{2}a(t_2-t_1)^2\)   ...(v)

Now if, x₁ =x₀ at t₁ = 0

x₂ =x and t₂ = t

v₁ = u at t₁ = 0

v₂ = v at t₂ = t

Eqn (v) reduces to \(x=ut+\frac{1}{2}at^2\).