Given the body is moving in one direction, thus the distance travelled by the body is equal to its displacement We know that the distance traveled in nth second is given by Sn = u+
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a(2n−1) Let D4 and D2 be the distances covered in the 4th and the 2nd seconds respectively. Then; D4 = 20 = u +
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a(2×4−1) ⇒ 20 = u+
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... (i) and D2 = 12 = u +
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a(2×2−1) ⇒ 12 = u+
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... (ii) Solving eqs. (i) and (ii), we get u = 6 m/s and a = 4 m∕s2 Using the second equation of motion : S = ut +
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at2, distance covered in 4 seconds after the 5th second is S9−S5 = [9u+