IIT JEE , AIEEE Forum - Physics , Mechanics

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Applying formula for distance traveled in nth second = u + (a/2)(2n-1)

When n = 3
u + (a/2)(2x3-1) = 4
2u + 5a = 8

When n = 5
u + (a/2)(2x5-1) = 12
2u + 9a = 24

From the above two equations we get a = 4 and u = -6

Distance traveled for 5 seconds = -6x5 + (4x5x5)/2 = 20 m

Distance traveled for 8 seconds = -6x8 + (4x8x8)/2 = 80 m

Hence distance traveled in next 3 seconds = 80 - 20 = 60 m

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