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A body covers 100 cm in first 2 seconds ...

A body covers 100 cm in first 2 seconds and 128 cm in the next four seconds moving with constant acceleration. Find the velocity of the body at the end of 8 sec?

Text Solution

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distance in first two seconds is
`s_(1) = ut_(1) + (1)/(2) at_(1)^(2)`
100 = `2u + (1)/(2) a(4) …….. (1)
Distance in (2 + 4) sec from starting point is
`s_(1) + s_(2) = u(t_(1) + t_(2)) + (1)/(2) a(t_(1) + t_(2))^(2)`
`228 = 6u + (1)/(2) a(36) .... (2)`
from eq (1) and (2)
we get a = -6 cm/`s^(2)`, sub a = -6 in eq - (1)
`rArr 100 = 2u - (1)/(2) xx 6 xx4`
`2u =112 u = 56 cm//s`
`v =u + at = 56-6xx8 rArr v = 8cm//s`
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