Question

a particle moving with a constant acceleration describe in the last second of its motion 9/25 of the whole distance.If its starts from rest, how long is the in motion and through what distance does it move if it describe in 6 cm in the first second?

Answer 1

body starts from rest so, initial velocity , u = 0
Let distance covered by body is d during t time.
we know, distance covered by nth second of a body is given by
Snth = u + 1/2a(2n - 1) , where a is acceleration and u is initial velocity
A/C to question,
last second body covered 9/25th of total distance ,
so, 9/25 of d = 0 + 1/2a(2t - 1)
⇒9d/25 = 1/2a(2t - 1)
⇒18d = 25a(2t - 1) --------(1)

Again, distance covered in first second is 6cm
S = 1/2at²
6cm = 1/2a × 1
a = 12 cm/s²

Now, Use formula
V² = u² + 2aS
v² = 0² + 2 × 12 × d [ S = d and a = 12cm/s²]
v² = 24d -----(2)

Again, v = u + at
v = 0 + 12t
v = 12t , put it in equation (2),
(12t)² = 24d
⇒144t² = 24d
⇒ 6t² = d -------(3), put it in equation (1),

18(6t²) = 25 × 12 (2t -1)
108t² = 600t - 300
27t² = 150t - 75
9t² = 50t - 25
9t² - 50t + 25 = 0
9t² - 45t -5t + 25 = 0
(9t - 5)(t - 5) = 0 ⇒ t = 5/9 and t = 5 but t ≠ 5/9

So, t = 5 sec
d = 6t² = 6 × 5² = 150cm

Hence, total distance covered by body is 150cm and time taken by body is 5sec

Answer 2

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