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.

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
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
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