can someone help me answer all of part C please? Thank you 1. An elevator shown below filled with passengers has a mass of 1700 kg. The elevator does motions (a)through (c) in succession.TyL.XImegFor each of the parts below draw a free body diagram of the elevator in your notebook for each of theparts (a) to (c). Draw the acceleration and velocity vectors in the boxes.mE(a) The elevator accelerates upward from rest at a rate of 0.752SaVfor 1.5 s. (c) The elevator decelerates at a rate of 0.35 for 2.75 s.2s²S(i) Newton's Law in the y-direction can be written as:Instruction: If a is the magnitude of the acceleration, pick "1" if the acceleration is upwards, pick"-1" if the acceleration is downwards, and pick "0" if there is no acceleration.ΣF₁=T-m₂g=xm a(ii) Calculate the tension in the cable supporting the elevator.T=✓N(iii) How high has the elevator moved during this time?Ay=✔m(d) What is the total distance the elevator has moved up?Ay net =Reflect: In which case, is the tension in the chord pulling the elevator the greatest? Why? Does thetension always equal the weight of the elevator?✓m

c) Given that Mass of the elevator(me) =1700kg Elevator decelerates at a rate of (a) =-0.35m/s2 The time taken is (t) =2.75s i) So, We can write T-meg =-mea Free body diagram isii) Tension in the cable is T= mg-ma T =me(g-a) =1700(9.81-0.35) =16082N iii) How high the elevator is moved is given by s=ut+(1/2)at2 s =ut-(1/2)at2 We know that v=u+at. Final velocity v=0 after reaching to a height h a =(v-u/t) Then the initial velocity is u =- at =-(-0.35)(2.75) =0.9625m/s2 The height of the elevator is s = (0.9625)(2.75)-(0.5)(0.35)(2.75)2 =2.6468-1.3234 =1.3234m iv) Now for the total distance we have add all the distances in different subparts we got in (a),(b) & (c)