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