A2000kg coaster is released 40 meters above the ground on a frictionless track. what is the approximate speed at ground level?

At the top, the coaster's gravitational potential energy is

     (mass) x (gravity) x (height) =

   (2,000kg) x (9.8m/s²) x (40m) = 784,000 joules.

If the track is frictionless, then its speed at the bottom is the same
as if it were simply dropped.  At ground level, ALL of its potential
energy has changed into kinetic energy.

                          Kinetic energy = (1/2) x (mass) x (speed)²

We know how much potential energy it had before it dropped,
and all of that energy is kinetic at the borrom:

                           784,000 joules = (1/2) (2,000kg) x (speed)²

Divide each side by 1,000:   784 = (speed)²

                             Speed= √784 = 28 meters per second

(That's about 62.6 miles per hour.)

Is a  vector quantity  that refers to "the rate at which an object changes its position." imagine a person moving rapidly - one step forward and one step back - always returning to the original starting position. while this might result in a frenzy of activity, it would result in a zero velocity. because the person always returns to the original position, the motion would never result in a change in position. since velocity is defined as the rate at which the position changes, this motion results in zero velocity. if a person in motion wishes to maximize their velocity, then that person must make every effort to maximize the amount that they are displaced from their original position. every step must go into moving that person further from where he or she started. for certain, the person should never change directions and begin to return to the starting position.