The distance travelled by the particle starting from rest and moving with an acceleration 4/3 m/s in the third second is
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Answered by
7
Answer:
S in nth second =u+a/2 (2n-1)
By substituting the above given values.
S=0+4/3÷2 ×(2×3-1)
S=2/3×5
S=10/3=3.3 (approx.)
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KingNipun1:
4/3÷2=2/3
Looser
But I am in 9
4/3÷2=4/ ÷ 1/2 =2/3
ab a ya samajh mae
Then 2/3 ×5 =10/3
3.3(approx.)
Ok
ab aaya samajh mae
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Answered by
3
Answer:
(4) 10/3m
Explanation:
i have provided the solution.
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