Question

A lift ascends from rest with uniform acceleration of 4 $m{s}^{-2}$, then it moves with uniform velocity and finally comes to rest with a uniform retardation of 4 $m{s}^{-2}$. If the total distance covered during ascending by the lift is 28 m and the total time for ascending 8 s, respectively, then find the time for which the lift moves with uniform velocity. Also, find its uniform velocity. Find the time for acceleration, retardation and uniform velocity.

Answer 1

Let the time for acceleration, retardation and uniform velocity be $t_1,t_3,t_2$ respectively.

Given,

$\frac{V_0}{t_1}=\frac{V_0}{t_3}=4$

$\therefore t_1=t_3$ . . . (i)

$V_0=4t_1$ . . . (ii)

Area under the velocity-time graph is the distance (provided the velocity is positive throughout the motion. If not, then the area will give displacement, not distance)

$(\frac{1}{2}\times t_1\times V_0)+V_0{t}_2+(\frac{1}{2}\times t_3\times V_0)=28$

$\Rightarrow V_0(t_1+t_2)=28$ . . . . . . . . . . . . using (i)

$\Rightarrow 4t_1(t_1+t_2)=28$ . . . . . . . . . . . . using (ii)

$\Rightarrow t_1(t_1+t_2)=7$ . . . (iii)

Given, total time of ascending = $8sec$

So $t_1+t_2+t_3=8$

$\Rightarrow t_1+t_2+t_1=8$ . . . . . . . . . . . . using (i)

$t_2=8-2t_1$ . . . (iv)

Puting (iv) this in (iii)

$t_1(t_1+8-2t_1)=7$

$t_1(8-t_1)=7$

$t_1^2-8t_1+7=0$

$t_1^2-7t_1-t_1+7=0$

$t_1(t_1-7)-1(t_1-7)=0$

$t_1=1or7$

$t_2=8-2t_1$

$t_1=1s$

$t_2=6s$

We rejected $t_1=7$ because it gives negative value of $t_2$ which is not acceptable, since time must be positive.

So,

$V_0=4t_1=4m/s$

From $t=0$ to $t=1s\to$ Acceleration

From $t=1s$ to $t=7s\to$ uniform velocity

From $t=7s$ to $t=8s\to$ retardation