Redwood
Graduate Poster
- Joined
- May 9, 2012
- Messages
- 1,557
Quick point, for the benefit of everyone, except FF, who couldn't possibly understand:
In kinematics, we have velocity, defined as change of position over time = dx/dt.
Then we have acceleration, change of velocity over time, or in differential calculus, the time derivative = dv/dt = dx/dt2. For an object at constant velocity, dv/dt = 0.
(I hope that none of this is going over anyone's head.
)
But we can also have a change in acceleration (which may have a name somewhere, but I don't know what it is) = da/dt = dv/dt2 = dx/dt3. An object falling in a vacuum has constant acceleration, so da/dt = 0.
But it's perfectly possible to have a changing rate of acceleration, as when one varies the thrust in a rocket; in which case da/dt =/= 0, and may be >0 or<0.
When you drop a brick, da/dt=0, but when it breaks through a piece of tissue paper, da/dt goes from zero to negative. This does not mean that the brick decelerates (dv/dt=negative). It merely experiences an interval where its acceleration was less than g.
In kinematics, we have velocity, defined as change of position over time = dx/dt.
Then we have acceleration, change of velocity over time, or in differential calculus, the time derivative = dv/dt = dx/dt2. For an object at constant velocity, dv/dt = 0.
(I hope that none of this is going over anyone's head.
But we can also have a change in acceleration (which may have a name somewhere, but I don't know what it is) = da/dt = dv/dt2 = dx/dt3. An object falling in a vacuum has constant acceleration, so da/dt = 0.
But it's perfectly possible to have a changing rate of acceleration, as when one varies the thrust in a rocket; in which case da/dt =/= 0, and may be >0 or<0.
When you drop a brick, da/dt=0, but when it breaks through a piece of tissue paper, da/dt goes from zero to negative. This does not mean that the brick decelerates (dv/dt=negative). It merely experiences an interval where its acceleration was less than g.

