1.

A lift starts from rest. Its acceleration is plotted against time. When it comes to rest its height above its starting point is A. `20m`B. `64m`C. `32m`D. `128m`

Answer» Correct Answer - D
At `t=4 sec. V=0+(4)(4)=16m//sec.`
At `t=8sec. V=16m//sec`.
At `t=12 sec, V=16-4(12-8)=0`
For `0` to `4 sec,s_(1)=(1)/(2)at^(2)=(1)/(2)(4)(4)^(2)=32m`
For 4 to `8 sec,s_(2)=16(8-4)=64m`
For `8` to `12 sec,s_(3)=16(4)-(1)/(2)(4)(4)^(2)=32m`
So `s_(1)+s_(2)+s_(3)=32+64+32=128m`
Alter`:` Draw `v-t` graph
Area of `v-t` graph `=` displacement.


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