A particle, after starting from rest experiences constant acceleration for 20seconds. If it covers a distance of S1 in first 10seconds and distance S2 in next 10seconds, then
A
S2=S1/2
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B
S2=S1
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C
S2=2S1
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D
S2=3S1
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Solution
The correct option is DS2=3S1 Let the constant acceleration be a
Given, u=0
For first 10seconds S1=ut+12at2 S1=0+12a(10)2 S1=50a …….(1)
Total displacement in 20seconds S=0+12a(20)2=200a ⇒S2=S–S1=150a.....(2)
From (1) and (2) S2S1=150a50a⇒S2=3S1