A body travels 2m in the 2nd seconds and 6m in the next four seconds . what will be the distance travelled in the 9th second.

Let the initial velocity of the body be ‘u’. Its acceleration be ‘a’.

It travels 2 m in the 2^nd second.

Using, S_n = u + (a/2)(2n – 1)

=> 2 = u + (a/2)(2×2 – 1)

=> 2 = u + 3a/2 ………………(1)

In the next 4 seconds it travels 6 m.

So, 6 m = distance traveled in 6 s – distance traveled in first 2 s

=> 6 = 6u + ½ a(6²) – 2u – ½ a(2²)

=> 6 = 4u + 16a

=> 3 = 2u + 8a ………………(2)

Solving (1) and (2) we get,

u = 2.3 m/s, a = -0.2 m/s²

So, distance traveled in the 9^th second is,

S₉ = u + (a/2)(2n – 1)

=> S₉ = 2.3 – (0.2/2)(2×9 – 1)

=> S₉ = 2.3 – (0.1)(17)

=> S₉ = 0.6 m