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?

Explanation

`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 =-7` in `eq- (1)`
`implies100=2u-(1)/(2)xx6xx4`
`2u=112" "u=56(cm)/(s)`
`v=u+at=56-6xx8implies v=8 cm//s`