A lift (elevator) starts from rest at the ground floor and moves vertically as follows:
• It accelerates upward uniformly for 3s until it attains a speed of 6ms−1.
• It then continues at this constant speed for the next 4s.
• Finally, it is brought to rest with a uniform acceleration of magnitude 2ms−2.
Take the upward direction as positive.
(i) State the sign and magnitude of the acceleration in each phase of motion.
(ii) Draw a labelled velocity–time sketch for the complete journey (no scale required).
(iii) Calculate the total distance moved by the lift from the instant it starts until it comes to rest.
(iv) If the downward direction were instead taken as positive, what would be the signs of the accelerations found in part (i) and the numerical value of the total displacement? Explain your reasoning.