- Good morning.
Today we are going to review integrals in kinematics,
as a part of the AP Physics C: Mechanics curriculum.
♫ Flipping Physics ♫
The equation for instantaneous acceleration is:
acceleration equals the derivative of velocity,
with respect to time.
And remember, every derivative can be rearranged
to make an integral, or an anti-derivative.
And we can do that like this:
we multiply both sides by dv, and then take
the definite integral of both sides.
Definite integral just means the limits are defined.
For example, on the left hand side,
when we take the integral with respect to velocity,
the limits are from velocity-initial, to velocity-final,
and on the right hand side, when we take
the integral with respect to time,
the limits are from time-initial, to time-final.
The integral of dv is just velocity,
from velocity-initial to velocity-final,
which equals velocity-final minus velocity-initial,
which equals the change in velocity,
in other words, change in velocity equals
the definite integral of acceleration
with respect to time.
Billy, the derivative represents the slope of a function,
what does the integral, or the anti-derivative represent?
- The integral is the area under the curve,
and the area under the curve specifically means
the area between the curve and the horizontal axis,
where the area above the horizontal axis is positive,
and the area below the horizontal axis is negative.
- Correct.
So the area between the acceleration curve,
and the time axis is the change
in velocity of the object.
And now, we can do the same thing
we just did with acceleration,
with velocity.
Starting with velocity equals the derivative of
position as a function of time,
we can arrive at the fact that
displacement equals the definite integral
of velocity with respect to time.
And that's highly redundant with
what we did with acceleration,
so I'm not gonna walk my way all the way through that.
Okay, let's move onto graphs of position, velocity,
and acceleration as a function of time.
Velocity is the derivative of position with respect to time.
Therefore the slope of a
position as a function of time graph is velocity.
Acceleration is the derivative of velocity with respect
to time, therefore the slope of a
velocity as a function of time graph is acceleration.
The definite integral of acceleration with respect to time
is change in velocity, therefore the area
under an acceleration as a function of time graph
is change in velocity.
The definite integral of a velocity with respect to time
is change in position, or displacement,
therefore the area under a velocity
as a function of time graph is displacement.
- Mr. P, what are these graphs for?
- [Mr. P] I don't know, what motion
do you all think these graphs represent?
- It looks like the acceleration is constant,
at just less than 10 meters per second squared,
so this is an object in freefall.
- The initial velocity is positive,
and the final velocity is negative,
and in the middle the velocity is zero.
- So these graphs represent the motion
of an object in freefall,
which was initially thrown upward?
- [Mr. P] Yes, that is correct.
If we assume the acceleration of an object
is constant as a function of time,
we can use the definition of instantaneous acceleration,
and indefinite integrals to derive
two of the uniformly accelerated motion equations.
We can rearrange the equation for instantaneous acceleration
to get dv equals acceleration dt,
and then we can take the indefinite integral of both sides.
The indefinite integral of just dv is
the velocity as a function of time,
and the indefinite integral of
acceleration with respect to time,
is acceleration times time, plus a constant, C.
In order to solve for the constant, C,
we can substitute an initial value of 0 in for time.
In other words, the velocity initial,
the velocity at time, t equals 0,
is equal to the acceleration times the 0,
plus the constant, C.
In other words, that constant C,
is equal to the velocity initial.
Substituting in the velocity initial,
for the constant, C, gives us
one of our uniformly accelerated motion equations:
Velocity final, or the velocity as a function of time,
equals the velocity initial, plus acceleration times time.
Bo, could you please do what we just did,
with the equation for instantaneous velocity?
- Uh, well, um ...
We know instantaneous velocity is the derivative
of position with respect to time,
multiply by dt to get dx equals v dt,
integrate both sides to get the integral of dx
equals the integral of velocity with respect to time.
The integral of dx is just x as a function of time,
but I don't know what to do with the integral
of velocity with respect to time.
In uniformly accelerated motion,
the acceleration is constant,
but the velocity changes with respect to time,
so I don't know what to do with that.
- We just derived the equation for velocity
as a function of time in
uniformly accelerated motion, so ...
- Oh yeah, so, we can substitute in for velocity
the equation velocity initial plus acceleration times time.
The integral of that with respect to time is
velocity times time, plus 1/2 times acceleration,
times time squared, plus a constant, C.
- And you can find the constant by letting time equal 0.
- Yeah, use the initial condition of time equals 0,
so that the position at time, t, equals zero,
which equals velocity times 0,
plus 1/2 times 0 squared, plus C.
That shows the initial position is the constant, C,
and therefore, the position as a function of time
equals the position initial,
plus velocity initial times time,
plus 1/2 times acceleration times time squared,
which is one of the uniformly accelerated motion equations.
- Very nice, that completes my review of
integrals in kinematics.
Next, feel free to enjoy my momentum and impulse review
for AP Physics C: Mechanics,
or you can visit my AP Physics C review webpage.
Thank you very much for learning with me today,
I enjoyed learning with you.