A body starts from rest and moves with a constant acceleration. It travels a distance s1 in the first 10s and distance s2 in the next 10s. Find the relation between s1 and s2. Share with your friends Share 0 Shivam Kumar answered this Dear Student, s=ut+12at2let acceleration is a,s1=0×10+12a102=50a ...........(i)And,s1+s2=0×20+12a202=200a ..........(ii)Dividing (ii) by (i),s1+s2s1=200a50a1+s2s1=4s2=3s1 Regards, 0 View Full Answer Yash Birje Harsh Birje answered this U=0 S1=1/2a(10)^2 S1=50a.... (1) S1+S2=1/2(20)^2 S1+S2=200a Subs (1), 50a+S2=200a S2=150a.... (2) (1)/(2) S1/S2=50a/150a S1/S2=1/3 Answer : S2=3S1 2