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```
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

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one year ago

Aman
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)s=16m.After this the lift moves with constant velocityTo 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
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one year ago
Khimraj
3007 Points
```							At 4s u=at=8m/ss1= ½*at2 = 16mfrom 4s to 8sa = 0, v= 8m/ss2 = vt = 8*4 = 32from 8s to 12ss3 = s1 = 16mtotal s1+s2+s3 = 64m
```
one year ago
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