The phrase *freely falling* really means *freely moving*, that is nothing is pushing or pulling the object while it's moving. So the object is *freely falling* even when it's *rising*! Like all things Physics has its own special terminology and I'm afraid you just have to get used to it.

Re the acceleration: the acceleration is the change of velocity with time. If I throw a stone upwards at 98.1 m/s then (ignoring air resistance) in the first second it slows by 9.81 m/s to 88.29 m/s, in the second second it slows by 9.81 m/s to 78.48 m/s and so on until after 10 seconds its velocity has slowed to zero. Each second it slows by the same 9.81 m/s, so the acceleration has the constant value of -9.81 m/sec$^2$ (the - sign means the acceleration is downwards).

Remember that velocity and acceleration are **vectors**, they have a *magnitude* **and** *direction*. Speed, on the other hand, is a *scalar* quantity which only has *magnitude*.

If you are moving (in a car) towards the right and apply the breaks, you are *decelerating* (slowing down)--your acceleration is directed towards the left while your velocity ("speed" in the diagram) is pointed towards the right (and is getting smaller):

(source)

So yes, the negative sign indicates a slowing down. A negative velocity would mean that the object is now moving in the opposite direction it was initially (or what you call "positive" direction, we could have made left the positive direction in the previous example).

The magnitude of your (negative) acceleration seems pretty high (though reasonable order of magnitude as to what I get below) given the high rate of speed initially ($v_1=150\,\rm m/s$), the zero final speed, and the extremely short stopping distance ($d=0.5\,\rm m$). You can use one of your kinematics equations,
$$
v_2^2=v_1^2+2a\Delta x,
$$
to get that $a=-22,500\,\rm m/s^2$, which is about 1/4 of what you have. Not sure where your mistake is, but I suspect it's in the calculation of the time (which doesn't seem necessary to me).

## Best Answer

The problem doesn't say the train "is slowing down at a constant velocity." That doesn't even make sense --- either it's slowing down, and the velocity is not constant, or it's not slowing down, and the velocity is constant. It can't be both accelerating and at a constant velocity, since the very definition of acceleration is change in velocity with respect to time.

Imagine you are driving in your car, dead north, which we'll call the positive direction. If you slam on the brakes, which direction will the acceleration be in? Since the car is slowing down, your acceleration will be in the South/negative direction.

The brakes are decreasing the car's velocity, which is the same thing as increasing it in the negative direction. Thus the sign of acceleration is negative.However, if you have pushed the accelerator to the floor, that would be a positive acceleration, since it is accelerating your car in its direction of positive velocity.

The same concept is true of the train here. Since it is slowing down over some frame of time, and it is moving in the positive direction, the acceleration will be negative. Since we're talking about tangential components, the fact that the train is in rotational motion isn't even relevant.