Question

A particle moving with constant acceleration describes in the last second of motion 9/ 25th whole distance . if it starts from rest how long is the particle in motion of and through what distance does it move if it describes 6 cm in 1 st sec. Plz answer

Answer 1

Explanation:

Here

u = 0

Let total distance covered be d

Let the particle move a total of k seconds

We have displacement at nth second

Sn = 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