Q.
lift performs the first part of ascent with uniform acceleration a and the remainder with uniform retardation 2a. The lift starts from rest and finally comes to rest. If t is the time of ascent. Find the height ascended by lift.
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a
at24
b
at23
c
at22
d
at28
answer is B.
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Detailed Solution
Let OAB be the velocity-time graph of the lift. The ordinate at A (i.e., AM) represents maximum velocity.Total distance travelled = Area of the ΔOAB = 12×OB×AMAM = v, OM = t1, t1+t2= OB = t, MB = t2 ∴ ΔOAB = 12×tv = hOr vt = 2h ..(i)Now vt1 = a or t1 = va .....(ii)and vt2 = 2a or t2 = v2a .....(iii)Adding (ii) and (iii)t = t1+t2 = va+v2a = 3v2a = 32a×2htor at2 = 3h ⇒ h = at23
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