A particle moves with uniform acceleration along a straight line from rest. The percentage increase in displacement during sixth second compared to that in fifth second is about

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Q: 7 (CDS-II/2018)
A particle moves with uniform acceleration along a straight line from rest. The percentage increase in displacement during sixth second compared to that in fifth second is about

question_subject: 

Science

question_exam: 

CDS-II

stats: 

0,9,17,8,9,8,1

keywords: 

{'uniform acceleration': [0, 0, 0, 5], 'particle moves': [0, 0, 0, 1], 'displacement': [0, 1, 0, 3], 'percentage increase': [0, 0, 0, 3], 'sixth second': [0, 0, 0, 1], 'fifth second': [0, 0, 0, 1], 'straight line': [1, 0, 3, 15]}

The problem states that a particle is moving with uniform acceleration along a straight line, starting from rest. We are asked to find the percentage increase in displacement during the sixth second compared to that in the fifth second.

When a particle moves with uniform acceleration, its displacement is given by the formula:

S = ut + (1/2)at^2, where S is the displacement, u is the initial velocity (which is 0 in this case), a is the acceleration, and t is the time.

To compare the displacements in the fifth and sixth seconds, we need to find the displacement at the end of each second. Let`s represent the displacement at the end of the fifth second as S5 and the displacement at the end of the sixth second as S6.

Using the formula, we have:

S5 = (1/2)a(5)^2 = 12.5a

S6 = (1/2)a(6)^2 = 18a

To find the percentage increase in displacement, we need to calculate the difference between S6 and S5 as a percentage of S5:

Percentage increase = [(S6 - S5)/S5] * 100

= [(18a - 12.5a)/(12