Question

a particle moves with constant acceleration along a straight line starting from rest. the percentage increase in its displacement during the 4th second compared to that in the 3rd second is? (ans : 40%)

2

Answer 1

Question:-

a particle moves with constant acceleration along a straight line starting from rest. the percentage increase in its displacement during the 4th second compared to that in the 3rd second is? (ans : 40%)

ANSWER:-

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=25a27a−