A body covers a distance 4m in 3rd sec and 12m in 5th sec. If the motion is uniformly accelerated, calculate initial velocity of the body.
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Let initial velocity be u and acceleration be a
sn = u + ½a(2n-1)
a/q
u + ½a(2×3-1) = 4 ---1.
u + ½a(2×5-1) = 12 ---2.
Eqn. 2 – eqn. 1
=> ½a(2×5-1) - ½a(2×3-1) = 8
=> 4a = 16
=> a = 4 ms-2
By 1.
u + ½ ×4×5 = 4
=> u = -6 ms-1
Total distance covered 8 s
= -6×8 + ½×4×(8)2
= 80 m
Distance covered in 5th second
= -6×5 + ½×4×(5)2
= 20 m
So distance covered in next 3 second after 5th second will be
80 – 20 = 60 m
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