Question

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

• A. $v=4m/s,t=6secs$
• B. $v=5m/s,t=9secs$
• C. $v=2m/s,t=8secs$
• D. $v=6m/s,t=5secs$

Answer 1

The correct option is A $v=4m/s,t=6secs$

There are $3$ cases

Case 1 : ascending with uniform accrltn

initial velocity $u=0m/s$ , a= 4m/s² , time $=t₁sec$ , final velocity $=vm/s$

distance covered in t₁ = s₁ m

$v₁=0+4t₁$ $,$ $2xs₁x4+0²=v²\Rightarrow s₁=v²/8$

case 2 : moving with uniform velocity $v$

since uniform velocity therefore no acceleration

initial velocity$=vm/s$ $,$ $a=0m/s²$ $,$ time $=tsec.$ distance in$t=s₂$ m final velocity = v m/s

$s₂=vt$

case3 : ascending with retardation

initial velocity $=vm/s$ , time taken $=t₂sec$ $,$ distance travelled $=s₃m$

accleration $=-4m/s²$ final velocity $=0m/s$

$2x-4xs₃+v²=0²$ $\Rightarrow s₃=v²/8$ $,$ $0=v-4t₂\Rightarrow t₂=v/4$

now $A/Q$

total time $=t₁+t+t₂=8\Rightarrow v/4+t+v/4$ $\Rightarrow t=8-v/2$..........$\left(1\right)$

total distance $=s₁+s₂+s₃=28$ $\Rightarrow v²/8+v²/8+vt=28\Rightarrow v²/4+4vt=112$

$\Rightarrow$substituting value of t from (1) gives quadratic equation in terms of v

$v²-32v+112=0$

on solving we get $v=4m/s$ and $v=28m/s$

$t=8-4/2=6sec$ and $t=8-28/2=-6sec$as time can't be negative

true ans is $v=4m/s$ and $t=6secs$

Hence

option $\left(A\right)$ is correct answer.