Question

a(m/s) F1. An elevator is on the 20° floor of a 40-floor building. At time 1-0, the elevator is at rest and the doors have just finished closing A student on the elevator has a smartphone with an app that can measure vertical acceleration. The student takes data for the 10 seconds starting at time the acceleration data is shown in the graph on the right. Let positive values represent upward acceleration (a) Qualitatively and quantitatively describe the motion of the elevator at the following times i Timer = 5 seconds li Timer = 10 seconds (b) At time 1 = 10 seconds, is the elevator above, below, or at its initial position? Explain your reasoning (e) On the grids below, draw graphs representing the velocity v and height y of the elevator as functions of time during the 10-second interval Assume that the initial height of the elevator is y = y. Be sure that your graphs have proper shapes and slopes during the appropriate times 343633 930 (3) 8 9 10 (8)

Answer 1

(a) at t= 2 see from t=0 sec the elevator is at rest so, no acceleration no velocity. from t=2 to t= u see there is an upward acceleration of 3 m/s2. So velocity increases. from t= a see to t= 6 See the acceleration is Zero means the lift is in Constant motion upward. from t=6 See to += 8 see there is an deacceleration of 3 m/s² so velocity decreases. from t=8 see to t= to see then there is no acceleration and velocity, the lift is at rest, (1) So at t= 5 see the lift is at const. motion in upwara where at t=10 the lif is at rest (b) From at t= 10 see analysis given the elevator above it is clear that above of its inital position

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