A lift ascends with a constant acceleration of 2 \mathrm{ms}^{-2}2ms −2 . Then it moves with a constant velocity and finally stops under a constant retardation of 2 \mathrm{ms}^{-2}2ms −2 . If the total distance ascended is 18m18m and the total time taken is 10 s, then the time(in s) taken to move with constant velocity i
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PHYSICS
A lift ascends from rest with uniform acceleration of 4m/s2,then it moves with uniform velocity and finally comes to rest with uniform retardation of 4m/s2. If the total distance covered during ascending is 28mand 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
December 20, 2019Prince Guda
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ANSWER
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⇒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 int=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 ⇒s₃=v²/8 , 0