A lift performs the first part of its ascent | vedclass

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(B) Let $t_1$ be the time of acceleration and $t_2$ be the time of retardation. The maximum velocity reached is $v = a t_1$.Since the lift comes to re

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$\frac{a t^2}{3}$

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(B) Let $t_1$ be the time of acceleration and $t_2$ be the time of retardation. The maximum velocity reached is $v = a t_1$.Since the lift comes to rest,$v = (2a) t_2$. Thus,$a t_1 = 2a t_2$,which implies $t_1 = 2 t_2$.Given the total time $t = t_1 + t_2$,we substitute $t_1 = 2 t_2$ to get $t = 2 t_2 + t_2 = 3 t_2$. Therefore,$t_2 = \frac{t}{3}$ and $t_1 = \frac{2t}{3}$.The total height $h$ is the sum of the distances covered during acceleration and retardation:$h = \frac{1}{2} a t_1^2 + (v t_2 - \frac{1}{2} (2a) t_2^2)$Substituting $v = a t_1 = 2a t_2$:$h = \frac{1}{2} a (2 t_2)^2 + (2a t_2 \cdot t_2 - a t_2^2) = 2 a t_2^2 + a t_2^2 = 3 a t_2^2$.Substituting $t_2 = \frac{t}{3}$:$h = 3 a (\frac{t}{3})^2 = 3 a (\frac{t^2}{9}) = \frac{a t^2}{3}$.

$A$ lift performs the first part of its ascent with uniform acceleration $a$ and the remaining with uniform retardation $2a$. If $t$ is the total time of ascent,find the total height of the shaft.

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