an
72. A body "A starts form rest with
acceleration a, After 28, another body B
starts from rest with an acceleration
a, if they travel equal distances
in the 5th Second after the starter of 'A.
the ratio of a, oda is equal to-
(6) 5:9 (6) %os (d)s : 7 7
then
9.5
Answers
Answered by
1
Answer:
5:9
Explanation:
Initial speed u=0
Initial speed u=0Distance covered in nth second after starting from rest S=21(2n−1)
Initial speed u=0Distance covered in nth second after starting from rest S=21(2n−1)For A : n=5
Initial speed u=0Distance covered in nth second after starting from rest S=21(2n−1)For A : n=5So, SA=2a1(2×5−1)=29a1
Initial speed u=0Distance covered in nth second after starting from rest S=21(2n−1)For A : n=5So, SA=2a1(2×5−1)=29a1For B : n=3
Initial speed u=0Distance covered in nth second after starting from rest S=21(2n−1)For A : n=5So, SA=2a1(2×5−1)=29a1For B : n=3So, SB=2a2(2×3−1)=25a2
Initial speed u=0Distance covered in nth second after starting from rest S=21(2n−1)For A : n=5So, SA=2a1(2×5−1)=29a1For B : n=3So, SB=2a2(2×3−1)=25a2But SA=SB
Initial speed u=0Distance covered in nth second after starting from rest S=21(2n−1)For A : n=5So, SA=2a1(2×5−1)=29a1For B : n=3So, SB=2a2(2×3−1)=25a2But SA=SBOr 29a1=25a2
Initial speed u=0Distance covered in nth second after starting from rest S=21(2n−1)For A : n=5So, SA=2a1(2×5−1)=29a1For B : n=3So, SB=2a2(2×3−1)=25a2But SA=SBOr 29a1=25a2⟹ a1:a2=5:9.
Hope it was helpful for you.
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