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Grade: 12th pass


A lift starts from rest it's acc is plotted against time in grah when it comes to rest it's height above station point is

one year ago

Answers : (2)

24 Points
Intially,the lift is accelerated with 2m/s^2 for 4 seconds starting from rest.
So the distance travelled in first 4s is :
s=0*4+(1/2*2*16) using (s=u*t+1/2*a*t^2) .....(1)
After this the lift moves with constant velocity
To find velocity at t=4s use (v=u+a*t)
So v=0+2*4=8m/s.
Now this lift covers s=8*4+1/2*0*16=32m...(from eqn 1)
Now the lift further deaccelerate in next 4s .. 
Again use eqn (1) but notice that this time initial velocity is 8m/s .
Pls approve if u find this answer helpful
one year ago
3007 Points
At 4s u=at=8m/s
s1= ½*at2 = 16m
from 4s to 8s
a = 0, v= 8m/s
s2 = vt = 8*4 = 32
from 8s to 12s
s3 = s1 = 16m
total s1+s2+s3 = 64m
one year ago
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