Question

Particle moves with constant acceleration
along straight line starting from rest.
The percentage increase in its displacement
during the 4th second compared to that

in the 3rd second is-
(a) 33% (b) 40% (c) 66% (d) 77%​

Answer 1

Answer:

40%

Explanation:

The displacement of the particle in nth second          Sn=u+21a(2n−1)

where  all the symbols have their usual meaning.

Given :      u=0

⟹Sn=21a(2n−1)

Thus displacement in the 3rd second             S3=21a(2×3−1)=25a

Displacement in the 4th second             S4=21a(2×4−1)=27a

Percentage increase in the displacement     S3ΔS×100=S3S4−S3×100

  S3ΔS×100=25a