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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
Explanation
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, $S = ut + 1/2a t^{2}$ $12 = u \times 3 + \frac{1}{2} \times a \times 3^{2}$ or $24 = 6u + 9a$ Similarly, $42 = u \times 6 + \frac{1}{2} \times a \times 6^{2}$ or $42 = 6u + 18a$ On solving, we get $u = 1 m s^{-1}$.