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 a uniform retardation of 4m/s^2. If the total distance covered during ascending is 28m and total time for ascending is 8s, respectively, then find the time for which the lift moves with uniform velocity. Also find its uniform velocity.

Answer 1

here r 3 cases  

Case 1 : ascending with uniform accrltn  

initial velocity u = 0m/s  ,  a= 4m/s² ,   time = t₁ sec ,  final  velocity =v m/s 
distance covered in t₁ = s₁ m 
v₁ = 0 + 4t₁        ,        2xs₁x4 + 0² = v²  ⇒ s₁=v²/8  

case 2 :  moving with uniform velocity v  

since uniform velocity therefore no acceleration   

initial velocity = v m/s    ,    a = 0 m/s²  ,  time = t sec  ,  distance in t = s₂ m  final velocity = v m/s 

s₂ = vt    


case3 :  ascending with retardation  

initial velocity = v m/s  ,  time taken = t₂ sec  ,    distance 
initial velocity = v m/s  ,  time taken = t₂ sec  ,    distance travelled = s₃ m  
accleration = -4 m/s²  final velocity  = 0m/s 

2x -4 x s₃  + v² = 0² ⇒  s₃= v²/8  ,    0 = v -4t₂ ⇒ t₂ = v/4   

now A/Q 

total time = t₁ + t + t₂ = 8 ⇒ v/4 + t + v/4 ⇒ t = 8 - v/2..........(1) 

total distance = s₁+ s₂ + s₃ = 28 ⇒  v²/8 + v²/8 + vt = 28 ⇒ v²/4 + 4vt = 112 
⇒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 = 6 sec  and  t = 8-28/2 = -6 sec as time can't be negative  

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