A PARTICLE STARTS MOVING WITH ACCELERATION 2M/S .DISTANCE TRAVELLED BY IT IN 5TH HALF SECOND

S=ut +1\2at^2 Initial speed is 2, then given that acceleration is2 and there is also given that , we want to find the distance when 5th half second means 4 sec completed and half means 4.5 show that t= is equal to 4.5a=2,u=2,t=4.5Hence s=29.25

Formula for distance in nth second is s= vi+a/2(2n-1)Acceleration is 2 and initial velocity is 2 so for fifth second S=2+2/2(2(5)-1))S=11 mSo for fifth half sec S= 11/2=5.5m

a = 2So v = integration of 2 wth respect to t = 2t s = integration from 0 to 2.5(2t)dt = t^2 from 0 to 2.5 = 6.25m