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A lift ascends with a constant acceleration of 4 m s^(-2) then with a constant velocity v and finally stops under a constant retardation of 4 m s^(-2). If the total height ascended be 20 m and the total time taken is 6 s then the time during which the lift was movitng with a vclocity nu is |
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Answer» 2a `nu=u +atimplies nu=0+4t` `t_(1)` is the TIME for which lift ascends with constant velocity Total time `= 2t+t_(1)=6s` Area under velocity - time GRAPH gives the displacement `(1)/(2) (2t+2t_(1))4t =20` `(t+t_(1))t=5` PUTTING value of `t_(1)=6-2t` from (i ) in (ii) `(t+6-2t) t=5` (6-t) t=5 `t^(2)-6t+5=0` (t-1)(t-5) =0 t=1 s or t=5 s , since t=5 s is not possble So, `t_(1)=4 s` 
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