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\includegraphics[max width=\textwidth, alt={}, center]{a96ca3b4-6d35-4512-a0a1-3f28443fd051-06_661_1529_260_306} An elevator moves vertically, supported by a cable. The diagram shows a velocity-time graph which models the motion of the elevator. The graph consists of 7 straight line segments. The elevator accelerates upwards from rest to a speed of \(2 \mathrm {~m} \mathrm {~s} ^ { - 1 }\) over a period of 1.5 s and then travels at this speed for 4.5 s , before decelerating to rest over a period of 1 s . The elevator then remains at rest for 6 s , before accelerating to a speed of \(V \mathrm {~m} \mathrm {~s} ^ { - 1 }\) downwards over a period of 2 s . The elevator travels at this speed for a period of 5 s , before decelerating to rest over a period of 1.5 s .

1. Find the acceleration of the elevator during the first 1.5 s .
2. Given that the elevator starts and finishes its journey on the ground floor, find \(V\).
3. The combined weight of the elevator and passengers on its upward journey is 1500 kg . Assuming that there is no resistance to motion, find the tension in the elevator cable on its upward journey when the elevator is decelerating.