Question

In rising from rest to rest in 8 seconds, a lift accelerates uniformly to its maximum speed and then retards uniformly. The retardation is a third of the acceleration and the distance travelled is 20m. Find the acceleration, retardation and the maximum speed reached

Answer 1

Let the maximum speed attained by the lift = $v_{1}$ m/s

Let the upward acceleration of the lift = $a$ m/s²

Net upward acceleration = $(a-g)$ m/s²

Retardation = $\dfrac{1}{3}a$ m/s²

Net retardation = $-(\dfrac{1}{3}a+g)$ m/s²

Total time taken by the lift = $8$ s

Total distance travelled = $20$ m

During upward acceleration :

$v=u+at_{1}$

$\Rightarrow v_{1}=0+(a-g)t_{1}$

$\Rightarrow t_{1}=\dfrac{v_{1}}{a-g}$ (1)

$v^2-u^2=2as_{1}$

$\Rightarrow v_{1}^2-0=2(a-g)s_1$

$\Rightarrow s_1=\dfrac{v_1^2}{2(a-g)}$ (2)

During retardation :

$v=u+at_{2}$

$\Rightarrow 0=v_{1}-(\dfrac{1}{3}a+g)t_{2}$

$\Rightarrow t_{2}=\dfrac{3v_{1}}{a-3g}$ (3)

$v^2-u^2=2as_2$

$\Rightarrow 0-v_1^2=-2(\dfrac{1}{3}a+g)s_2$

$\Rightarrow s_2=\dfrac{3v_1^2}{2(a+3g)}$ (4)

Total time taken,

$t_{1}+t_{2}=8 s$

$\Rightarrow \dfrac{v_{1}}{a-g}+\dfrac{2v_{1}}{a-3g}=8$

$\Rightarrow v_{1}a=2a^2+4ag-6g^2$ (5)

Total Distance covered,

$s_1+s_2=20m$

$\Rightarrow \dfrac{v_1^2}{2(a-g)}+\dfrac{3v_1^2}{2(a+3g)}=20$

$\Rightarrow v_1^2a=10a^2+20ag-20g^2$ (6)

Multiplying eq.(5) by 5 and subtracting from eq.(6), we get

$\Rightarrow v_1^2a-5v_1a=0$

$\Rightarrow v_1a(v_1-5)=0$

Since $v_1a\cancel{=}0$,

$\Rightarrow v_1-5=0$

$\Rightarrow v_1=5ms^{-1}$

Similarly we can find out the value of acceleration by putting the value of $v_1$ in eq.(5)

$\Rightarrow a=10.45ms^{-2}$

Retardation $=-\dfrac{1}{3}a=-3.48ms^{-2}$