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PROFESSOR WALTER LEWIN: What now is the instantaneous acceleration a?
The instantaneous acceleration a is defined as the time
derivative of the velocity.
If I take a one-dimensional case-- which I always like to do first,
because then we can remove the vectors.
Because the signs then automatically take care of the directions.
So always start with a one-dimensional case.
So we have now v, for instance, equals 3t^2-5.
And so we would then have a equals 6t.
And this number could be positive, could be zero, could be negative.
In fact, if t equals 0, you will see that the acceleration is zero.
If t is larger than zero, for positive numbers, the acceleration is positive.
And for negative numbers of t, smaller than zero, the
acceleration is negative.
And so the signs here automatically take care of the direction.
For values lower than zero, times lower than zero, the acceleration is
in the minus x direction.
For values larger than zero the acceleration is in
the positive direction.
Wonderful advantage of one-dimensional situations.
Now let's turn to a more complicated situation, the
three-dimensional situation.
3D--
let the velocity vector be 3t x-roof minus 2t^2 y-roof plus 4z.
And distances are always in meters, and times always in seconds.
The components of the velocity in the x direction and in the y direction are
time dependent.
But this component in the z direction is not time dependent.
a, instantaneous value for a, equals dv / dt.
And that now equals 3x-roof minus 4t y-roof.
So the acceleration has a component in the x direction and has a component in
the y direction.
It has no component in the z direction.
If we take t=+3, then we would have that the acceleration, which is a
vector, at t equal 3, at a particular moment in time-- the instantaneous
acceleration at t=3 would be plus 3x-roof minus 12y-roof.
So a_x = +3 m / s^2 and a_y, the y component of the acceleration, would
be -12 m/s^2.
a_z would be zero, and the magnitude of a at this moment in time t=3 would
be the square root of 3 squared plus minus 12 squared
meters per second squared.
When the acceleration is constant, when it is independent of time, the
time averaged acceleration between any moments in time is exactly the same as
the instantaneous acceleration at any moment in time that you choose.
And I would like to revisit something that I did earlier whereby we had v_x
= 3t+4, v_y = -2t and v_z was plus +6.
We calculated the average acceleration between time t = 1 and time t = 4.
And what did we find?
We found 3x-roof minus 2y-roof.
And we picked these two times randomly.
Now let's calculate the instantaneous acceleration a, which is dv / dt.
a _ x = + 3.
a _ y = - 2.
a _ z = 0.
So what is the instantaneous acceleration at any moment in time?
That is plus 3x-roof minus 2y-roof.
And so you see that the instantaneous acceleration is independent of time,
so the acceleration is constant.
It doesn't change in time.
And so it shouldn't surprise you that when you calculate the time averaged
acceleration between any two points in time, any two that you can choose,
that you find exactly the same results.
We will often encounter situations in Newtonian mechanics whereby we have a
constant acceleration.
Whenever we deal with trajectories, we throw rockets, and we ignore air drag,
we get a nice parabola.
And if we call this the plus y direction, and this is called the x
direction, which is quite common--
we will have no acceleration in the x direction unless there's air drag.
And there will be an acceleration in the minus y direction, which is 10
meters per second squared.
So we will often deal in this course, with situations whereby we do have
constant acceleration.