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

We know that, snth=u+12a(2n-1)       s3rd=0+12a(2×3-1)=52a                       (For n = 3s)       srrd=0+12a(2×4-1)=72a                        (For n = 4s)So, the percentage increase =s4th-s3rds3rd×100           =72a-52a52a×100=2a252a×100=2×20=40%