Physics, asked by aarav9206, 1 year ago

A particle moving with constant acceleration describes 9/25th of the whole distance in the last second of.Its motion

Answers

Answered by poonambhatt213
5

The distance covered by particle is 150 cm and the particle is in motion for 5s.

Explanation:

Q: A Particle moving with constant acceleration describes in the last second of it's motion 9/25th 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, particle starts from rest, so u = 0

Suppose, total distance covered = d and  

total time taken by particle = k sec

=> Thus, displacement at nth second :

Sn = u + ½ a (2n-1)

[9/25]*d =  0 + ½ a (2k-1)

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

=> If it describes 6 cm in it's first second then,

6 = ½ a (2-1)

a = 6*2

a = 12  ...(2)

=> In addition,  v² – u² = 2ad

but u = 0

v² = 2×12×d  

= 24d  

= √(24d)

=> even, t = k

v = u + ak

√(24d) = 0 + 12k

24d = (12k)²

24d = 144 k²

d = 144 k² / 24

∴ d = 6k²  ...(3)

=> From eq(1) and eq(3), we get

108k² = 25×12×(2k-1)

108k² = 300 (2k - 1)

108k² = 600k - 300 = 0

108k² - 600k + 300 = 0  

4(27k² - 150k + 75) = 0

27k² - 150k + 75 = 0

=> either k = 5/9 s or 5s

but 5/9s is not possible as particle travels more than one sec)

 So, the particle moves for 5 seconds

Thus, the distance covered by it,

d = 6k²

= 6*5²

= 6*25

= 150 cm

Thus, the distance covered by particle is 150 cm and the particle is in motion for 5s.

Learn more:

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