A lift is going up. The total mass of the lift and the passengers is 1500kg. The variation in the speed of the lift is given by the graph.

(a) What will be the tension in the rope pulling the lift at time t equal to
(i) 1a, (ii) 6s, (iii)11s ?
(b) What will be the average velocity and the average acceleration during the course of the entire motion?

Explanation

From `t=0` to `2 s`, the lift has uniform acceleration
`a_(1)=(Delta v)/(Delta t)=(3.6-0)/(2-0)=1.8 ms^(-2)`
From `t=2 to 10s` the left has a constant speed and hence acceleration `a_(2)` during the period is `0m//s^(2)`
From `t=10s to 12 s`, the lift has a uniform acceleration. If `a_(3)` is the acceleration during the period , then
`a_(3)=(Delta v)/(Delta t)= (v_(2)-v_(1))/(t_(2)-t_(1))=(3.6-0)/(2-0)=1.8 ms^(-2)`
(a) Here `m=1500 kg, g=9.8 ms^(2)`
(i) At `t=1s, a=a_(1)=1.8 ms^(-2)`
Tension `T_(1)=m(g+a)=1500(10+1.8)=17700N`
(ii) At `t=6s, a=a_(2)=0`
Tension `T_(2)=m(g+a_(2))=mg=1500 xx 10 = 15000N`
At `t=11s, a=a_(3)=-1.8ms^(-2)`
Tension `T_(3)=m(g+a_(3))=1500(10-1.8)=12300N`.
(b) Height reached by the lift
=Area of enclosed by v-t curve and t-axis.
=Area of quadrilateral OABC`=36m`
average velocity =`("Total diplacment")/("Total time")=26/12=3 ms^(-1)`
Average acceleration = `("Net change in velocity")/("Total time")`
`=(0)/(12)=0 ms^(-2)`.