An elevator is stopped at the ground floor. It starts moving upwards at constant acceleration a>0 for 5 seconds. It then keeps a constant speed for 35 seconds. Finally, it slows down with an acceleration of the same magnitude (but opposite direction) −a , until it comes to a halt at the top floor. The top floor is 450 meters above the ground floor.

(a) What is the maximal speed v of the
elevator ? (in m/s)

(b) What is the acceleration a ? (in m/s2)

find acceleration: a=changev/time=35/5=7m/s^2

avg speed during acceleration, deacceleration= 17.5m/s

Now, how far does it go accelerating/deaccelerating? distance=17.5*5m=87.5m , and then again 87.5m slowing down.

so it must go 450-35 m at constant speed, for 35 seconds.

max speed=(450-175)/35 = you do it.