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2.Motion in Straight Line
normal
A lift performs the first part of its ascent with uniform acceleration $a$ and the remaining with uniform retardation $2a$. If $t$ is the time of ascent, find the depth of the shaft.
A
$\frac{a t^2}{4}$
B
$\frac{a t^2}{3}$
C
$\frac{a t^2}{2}$
D
$\frac{a t^2}{8}$
Solution
(b)
If $t_0$ is the time during acceleration, then $\frac{t_0}{2}$ will be the time during retardation.
Now,$t_0+\frac{t_0}{2}=t$
$\therefore \quad t_0=\frac{2 t}{3}$ and $\frac{t_0}{2}=\frac{t}{3}$
$t_0=\frac{2 t}{3}$ and $\frac{t_0}{2}=\frac{t}{3}$
$s=\frac{1}{2} a\left(\frac{2 t}{3}\right)^2+\frac{1}{2} \times 2 a \times\left(\frac{t}{3}\right)^2=\frac{a t^2}{3}$
Standard 11
Physics
