Question

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​

Answer 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.