(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
v2 - u2 = 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 v2 = 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 vav = 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, vav = \(\frac{v_1+v_2}{2}\) ...(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 \(x=ut+\frac{1}{2}at^2\).
