Setup: A rocket is launched from the ground with an upward acceleration of 12.5 m/s². After 3.5 seconds the engines of the rocket shutoff. How high will the rocket rise in the absence of air resistance.

Observations:

• Both before and after the engines shutoff the rock has a constant acceleration - 12.5 m/s² upwards before and 9.80 m/s² downward after. This is shown by the straight lines on the velocity vs. time diagram.
• The rocket continues to rise after the engines are shutoff because the rocket has an upward velocity at that moment even though its acceleration is downward, y = -g t +v_o
• The dots left behind represent the location of the rocket every .2 sec. They are not symmetric on the way up and down because the rocket does not reach its maximum altitude in a time which is exactly an integral multiply of exactly .2 sec. Even if this were the case the dots would be different when the rocket's height is less than the height when the rocket turns off.
• The rocket reaches its largest upward velocity at the moment the engines shutoff. 
• Note that the distance vs. time diagram represents two parabolas joined at the moment the rocket shuts off at 3.50 sec. The parabolas are inverted since one represents a positive acceleration on the way up and the negative acceleration during free-fall.
• The fact that the rocket makes a "perfect landing" show that we are working with models of reality and not reality itself.
• See Rising Rocket Problem - Rising Rocket IP Simulation