A particle starting with certain initial velocity and uniform acceleration covers a distance of $$12\,m$$  in first $$3$$ seconds and a distance of $$30\,m$$  in next $$3$$ seconds. The initial velocity of the particle is

Solution :
Let $$u$$ be the initial velocity that have to find and $$a$$ be the uniform acceleration of the particle.
For $$t = 3\,s,$$   distance travelled $$S = 12\,m$$   and for $$t = 3 + 3 = 6\,s$$    distance travelled $$S' = 12 + 30 = 42\,m$$
From, $$S = ut + \frac{1}{2}a{t^2}$$
$$\eqalign{ & 12 = u \times 3 + \frac{1}{2} \times a \times {3^2} \cr & {\text{or}}\,\,24 = 6u + 9a\,......\left( {\text{i}} \right) \cr} $$
Similarly, $$42 = u \times 6 + \frac{1}{2} \times a \times {6^2}$$
$${\text{or}}\,\,42 = 6u + 18a\,......\left( {{\text{ii}}} \right)$$
On solving, we get $$u = 1\,m{s^{ - 1}}$$