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 $$\(2 \mathrm{~m} \mathrm{~s}^{-1}\)$$ over a period of $$\(1.5 \mathrm{~s}\)$$ and then travels at this speed for $$\(4.5 \mathrm{~s}\)$$, before decelerating to rest over a period of $$\(1 \mathrm{~s}\)$$. The elevator then remains at rest for $$\(6 \mathrm{~s}\)$$, before accelerating to a speed of $$\(V \mathrm{~m} \mathrm{~s}^{-1}\)$$ downwards over a period of $$\(2 \mathrm{~s}\)$$. The elevator travels at this speed for a period of $$\(5 \mathrm{~s}\)$$, before decelerating to rest over a period of $$\(1.5 \mathrm{~s}\)$$ Given that the elevator starts and finishes its journey on the ground floor, find $$\(V\)$$. ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................

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