Question

If a particle moves with constant accleration and it describes its last second 9/25th of the whole distance.

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