PROFESSOR: 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 for us 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
squared minus 5.
And so we will then have a equals 6t.
And this number could be positive, could
be 0, could be negative.
In fact, if t equals 0, you will see that the
acceleration is 0.
If t is larger than 0 for positive numbers, the
acceleration is positive.
And for negative numbers of t, smaller then 0, the
acceleration is negative.
And so the signs here automatically
take care of the direction.
For values lower than 0 times lower than 0, the acceleration
is in the minus x-direction.
For values larger than 0, 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.
3-D.
Let the velocity vector be 3t x roof minus 2t
squared y roof plus 4z.
And distances are always in meters and
time's 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 3 x 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 equals plus 3, then we would have that the
acceleration, which is a vector at t equals 3 is at a
particular moment in time-- the instantaneous acceleration
at t equal 3 would be plus 3 x roof minus 12 y roof.
So a of x equals plus 3 meters per second squared.
And a of y, the y-component of the acceleration would be
minus 12 meters per second squared.
a of z would be 0 and the magnitude of a at this moment
in time t equal 3 would be the square root of 3 squared plus
minus 12 squared meters per seconds squared.
When the acceleration is constant, when it is
independent of the time, the time averaged acceleration
between any moment 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 of x 33 plus 4.
v of y equals minus 2t and v of z was plus 6.
We calculated the average acceleration between time t
equals 1 and time t equals 4.
And what did we find?
We found 3 x roof minus 2 y roof.
And we picked these two times randomly.
Now let's calculate the instantaneous acceleration a,
which is dv dt.
a of x equals plus 3, a of y equals minus 2,
a of z equals 0.
So what is the instantaneous acceleration at any moment in
time that is plus 3 x roof minus 2 y 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 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 a constant acceleration.