Verified answer

s = vt + (1/2)a t^2

You know the initial velocity is 0, acceleration is 2 m/s^2, and time is 2 s. Plug the numbers in and chug the results. For the second part, just increase the time to 4 s and chug again.

Homework?

s = s0 + V0 * t + 1/2 * a * t^2

We will have to make a lot of assumptions. Just because the trolley "can accelerate" at a constant acceleration does not mean it will. We are left to assume the acceleration is constant. We are not told whether the trolley starts at any velocity. We are left to assume the acceleration period starts with the trolley motionless. This counters the first assumption since an object at stand still will need to go from zero acceperation to some constant acceleration - it is impossible for that to be instantaneous. The equation I provided assumes constant mass, but the mass is changing, isn't it? The air leaves the trolley. Air has mass. We are not given enough information to calculate the effects of the mass flow rate. We must assume the mass of the expelled air is insignificant.

distance#1 = 1/2 (2 m/s^2) (2 s)^2 = 4 m

distance#2 = 1/2 (2 m/s^2) (2x2 s)^2 = 16 m

Those are the unthinking answers to the questions. Yet, the problem has a hidden complexity to it. "before running out of air"? The trolley is being propelled into a vacuum. If this trolley reaches the vacuum while at 2m/s^2 there is nothing to deaccelerate it. It will go on forever. Hence, the real answer is "to infiinity and beyond!"

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