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
11%
22%
33%
44%
Answer is Wrong!
Answer is Right!
This question was previously asked in
UPSC CDS-2 – 2018
Displacement in the fifth second (n=5): S₅ = a(5 – 1/2) = a(4.5).
Displacement in the sixth second (n=6): S₆ = a(6 – 1/2) = a(5.5).
The percentage increase in displacement during the sixth second compared to the fifth second is calculated as [(S₆ – S₅) / S₅] * 100%.
Percentage Increase = [(a(5.5) – a(4.5)) / a(4.5)] * 100% = [a(5.5 – 4.5) / a(4.5)] * 100% = [a(1) / a(4.5)] * 100% = (1 / 4.5) * 100% = (10 / 45) * 100% = (2 / 9) * 100% ≈ 0.2222 * 100% ≈ 22.22%.
The closest option is 22%.
– Uniform acceleration from rest means u=0.
– Percentage increase calculation involves finding the difference, dividing by the original value, and multiplying by 100.