A body describes 0.12 m in the 2nd second and 0.52 m in the 4th second of its motion. If the motion isuniformly accelerated, find how much distance will itcover in 5th second.

A body describes 0.12 m in the 2nd second and 0.52 m in the 4th seco

1.	A body describes 0.12 m in the 2nd second and 0.52 m in the 4th second of its motion. If the motion isuniformly accelerated, find how much distance will itcover in 5th second.
Answer» s n =u+a/2(2n−1)s 2 =12⇒12=u+a/2(2×2–1)⇒12=u+3a/2 ………. (1)s 4 =20⇒20=u+a/2(2×4–1)⇒20=u+7a/2 ………. (2)SUBTRACT equation (1) from (2)s=4a/2⇒a=4ms −2 12=u+3/2×4⇒u=6ms −1 Distance covered in the 4 seconds after 5 TH second,=S 9 –S 5 =u(9)+1/2a(9) 2 –[u(5)+1/2a(5) 2 ]=4u+a/2(81–25)=4×6+4/2×56=136m.

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A body describes 0.12 m in the 2nd second and 0.52 m in the 4th second of its motion. If the motion isuniformly accelerated, find how much distance will itcover in 5th second.​

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A body describes 0.12 m in the 2nd second and 0.52 m in the 4th second of its motion. If the motion isuniformly accelerated, find how much distance will itcover in 5th second.