A Particle moving with constant acceleration describes in the last second of it's motion 9/25^th of the whole distance. If it starts from rest how long is the particle in motion and through what distance does it move if it describes 6 cm in it's first second?

Here

u = 0

Let total distance covered be d

Let the particle move a total of k seconds

We have displacement at nth second

S_n = u + ½ a (2n-1)

A/Q

(9/25)×d =  0 + ½ a (2k-1)

=> 18d = 25a(2k-1) ---1.

Again in the first second it covers 6 cm thus

6 = ½ a (2-1)

=> a = 12 ---2.

Again

v² – u² = 2ad

u = 0

v² = 2×12×d = 24d  (by 1.)

=>v = √(24d)

Again we have, t = k

v = u + ak

√(24d) = 0 + 12k

=> 144k² = 24d

=>d = 6k²  ---3

From eqn. 1 and eqn. 3

108k² = 25×12×(2k-1)

=>108k² -600k + 300 = 0

=> 27k² – 150k+ 75 = 0

Either k = 5/9 s (not possible because particle travels more than 1 second)

Or k = 5 s

Thus particle moves for 5 seconds

It covers a distance

d = 6k²

=> d = 6×(5)² = 150 cm