A 600kg lifts starts from rest. It moves upward for 3.00 s with a constant acceleration until it reaches its final speed of 1.55 ms^-1 . calculate the force exerted by lift motor

We know that,

Force = mass× acceleration

We already have mass so let us calculate accleration experienced by it, using first equation of motion;

v = u + at1.55 = 0 + a(3)1.55 = 3a1.55/3 = a0.51 m/s² = a

Put this in force's formula;

f = maf = 600 × 0.51f = 306.00f = 306 N

Therefore, the force exerted by the lift is 306 N.

You need to be more specific.  i really don't know what a "motion map" is.you can draw a graph of an object's position, distance, displacement, speed,velocity, or acceleration versus time, or you can draw a graph of the object's x-position versus its y-position and not give any information about time at all.if the graph presents position or displacement against time,  then constant velocity is a straight line . . i.e. a line that has  constant slope.if the graph shows speed against time, then constant velocityis a horizontal piece of line.if the graph shows acceleration against time, then constant  velocity is a piece of line that lies on the x-axis . . it's value is zero.