We will discuss velocities
and acceleration. I'll start
with something simple. I have a motion of an object
along a straight line-- we'll call that
one-dimensional motion. And I'll tell you that
the object is here at time t1. At time t2, it's here. At time t3, it's there. At time t4, it's here and at time t5, it's back
where it was at t1. And here you see
the positions in x where it is located
at that moment in time. I will define this to be
the increasing value of x. It's my free choice,
but I've chosen this now. Now we will introduce what
we call the average velocity. I put a bar over it. That stands for average
between time t1 and time t2. That we define in physics as x at time t2
minus x at time t1 divided by t2 minus t1. That is our definition. In our case, because
of the way that I define the increasing value of x,
this is larger than 0. However, if I take the average
velocity between t1 and t5 that would be 0, because
they are at the same position so the upstairs is 0. If I had chosen t4 and t2-- average velocity
between time t2 and t4-- you would have seen
that that is negative because the upstairs
is negative. Notice that I haven't told you where I choose my zero
on my x axis. It's completely unimportant
for the average velocity. It makes no difference. However, if I had chosen this to be the direction
of increasing x then, of course,
the signs would flip. Then this would
have been negative and this would
have been positive. So the direction,
that you are free to choose determines the signs. The location where you put
your zero is not important but signs in physics do matter. Signs are important. Whether you owe me money
or I owe you money the difference
is only a minus sign but I think it's important
for you. Now I will give you
not only the positions-- as I did here on the x axis
at discrete moments in time-- but I'm going to tell you exactly where the object is
at any moment in time. Here you see an xt diagram so you see that at t1,
the object is at position xt1. This is the road of the object. This is the straight line,
where it's moving. It starts here and
it goes to this position. It goes to this one,
it comes back to t4 and it comes back here. I will tell you now
every moment in time in between. There it goes. Voilà. This is now information
that is way more. You have the information
at any moment in time. Notice that I now did choose
x = 0. I chose it somewhere here but I could have chosen it
at any other point-- for whatever follows you will see that it makes
no difference-- so I have chosen a zero point
so that I can make a graph. And now we will look
at the average velocity in a somewhat different way. Say I choose my time t2 and t3. I draw here now this line. And this angle I call alpha and this part here
I call delta x and this here is delta t. And so you could right now-- if you're careful
about your sign convention-- you could write down now
that the average velocity equals delta x
divided by delta t. But be careful. If the angle is positive--
I call this a positive angle-- then the average velocity
is positive but if I have a negative angle then the average velocity
would be negative. For instance, between t4 and t5,
if I draw this line then this angle here is negative and so the average velocity
between t4 and t5 is now negative. Again, if I had changed
the zero points you would have found
the same values for the average velocity. The only difference
would have been the position of the curve
in that plot. There is a very
big difference in physics between speed and velocity. The average velocity
between time t1 and t5 is zero but the average speed is not. The average speed is defined
as the distance traveled divided by the time that it
takes to travel that distance. Now, what is the distance
that the object traveled between time t1 and time t5? Well, the object started at
a position here on this x axis and then it went up,
reached the highest point here so I'll make a drawing
for you here. It reached the highest point
here, then it went down. And then when it went here it went up again and comes
down again and it's back. And in order to find
the average speed you would now have to know
exactly what this distance is add up this distance add up this distance
and this distance. And if that distance altogether
were, for instance, 300 meters and if the time between t1
and t5 were three seconds then the average speed would be 300 meters divided
by three seconds. That would be
100 meters per second so the average speed
would be 100 meters per second yet the average velocity
would be zero. If you look
at the location t3 and t2 and I bring t3 closer
and closer to t2 then this angle
of alpha will increase and I can go to the extreme that I bring t3
almost right at t2. The angle of alpha will then
be tangential to this point. This will then be
my angle of alpha. And now you will understand
how we define the instantaneous velocity
at time t which is different
from an average velocity between two time intervals. The instantaneous velocity, v--
and I pick a random time, t-- equals the limiting case for x measured at time t
plus delta t minus x measured at time t
divided by delta t and I do that for delta t--
goes to zero. So think of this
as being t3 and this as t2. I bring t3 closer and closer
and closer to t2 and the time between them
then goes to zero. And this is something
that you undoubtedly recognize. That's the first derivative
of the position versus time. And now comes an equation which is one of the very few that I want you to remember
in x... in 801: v equals dx/dt. This is one that you must
remember, not only in 801 but for the rest
of your time at MIT. And this could be larger than 0,
this could be 0 and this could be
smaller than 0. If the angle of alpha,
the tangential, is positive then it is a positive value. If it is negative, however,
when you're here then it is a negative velocity. And if the angle
of alpha is zero then it... the velocity is zero. So if we now look at this plot we can search for the times
that the velocity is zero so you have to look
for the derivative being zero. That means
the angle alpha being zero. Clearly, here the velocity
is zero. Right here,
at this turning point-- that means when the object
is here-- it is zero. When the object is here it is again zero
at this moment in time. Again, the angle is zero,
and it is again zero here. So those are the times
that the velocity is zero. What are the times
that the velocity is positive? Well, it's positive here. The velocity's positive here still positive, positive,
becomes negative negative, positive,
zero, negative. So that's the definition
of v, instantaneous velocity. What is the instantaneous speed? Well, speed is
not sign-sensitive. Suppose that the velocity here--
just... I call that v1-- suppose that was
plus 30 meters per second. I just grabbed this number
out of the blue. And suppose here, somewhere,
it was... I call that v2-- suppose that was
minus 100 meters per second. This is negative
and this is positive. Then we would have
to say, in physics-- whether you like it or not,
it's not very pleasing-- but you would have to say that this velocity
is lower than that one because minus 100
is lower than plus 30. But the speed, of course,
is higher here because the speed is
the magnitude of the velocity and is not sign-sensitive. So this has the highest speed,
of 100 meters per second and this has a lower speed but this has
the lowest velocity. It's just an algebraic game but very important
when you make your calculations. I have always wondered
what the average speed or the average velocity
is of a bullet. Now I want you to realize
I am not a fan of guns at all but it always intrigued me. How can I measure the average
speed of a bullet-- and I have discussed it
with some people here-- and we came up
with an easy way to do that. We have a wire, which goes
into the blackboard, wire I and we have
another wire that goes into the blackboards, wire II,
and the separation is D meters. We have to measure that. The set-up is here so this is wire number I
and this is wire number II. So you will see D
coming in like this, so I'll make this a I
and I'll make this a II. That's the way it's set up. And we fire the bullet,
which breaks this wire. At that moment, the timer starts and then it breaks this wire
and that's when the timer stops. Now, I told you
a measurement is meaningless without knowledge of the
uncertainty in your measurement. So there are
two uncertainties involved-- the distance
and the timing uncertainty. This distance
I will measure for you, D. I have here a large ruler. Here's one wire,
here's the other wire. I cannot do that
any better, really than maybe even half
a centimeter because the situation
is not all that stable-- I don't know what happens when
the bullet will hit the wire-- so I would say
it is 148½ centimeters but I cannot guarantee it to
better than half a centimeter-- 148½ plus or minus
0.5 centimeters. I want you to appreciate that this is a very small
percentage error. This is only five parts
out of 1,500. That is one out of 300, so that is only
a one-third percent error. That's very small-- that's
what we call the relative error. Then I ask myself the question-- I want to measure the accuracy
of the speed of the bullet to about two percent. That was my goal. How accurate
should I do the timing? Well, I had to make
an estimate very roughly how fast the speed
of the bullet is and-- I would think it is probably
lower than the speed of sound. The speed of sound
is 340 meters per second. I don't know
whether it's 200 or 300 but it's got to be somewhere
in that ballpark of the kind of bullets
that we have-- 200 or 300 meters per second. Let us assume that the speed
is 300 meters per second-- just a wild guess. Then it would take
5 milliseconds for this bullet to cross
from here to here. And if I want
to make a measurement to two percent accuracy I have to know this timing to about one-tenth
of a millisecond because one-tenth
of a millisecond is about two percent of five. So that sets the accuracy that I need to make
the time measurements. And so we do have a timer. It is about accurate to
about a tenth of a millisecond and so now I can measure
that time. So I am going to have here some time that we measure
plus or minus 0.1 and we'll do the whole thing
in milliseconds. But our final answer will be
in meters per second. All right, I always have
to think hard when I do this because when we deal with
bullets, that is no kid stuff and I... as I said I have really no experience
firing guns. This is the bolt. There we go. Here's the bolt. There we go. It's in place. Before I do that, I want
to check... check the circuits. I want to make sure
that the electronic circuit is properly working. You see the timing here, right? So I do a small test just to see whether
the circuit is working. Yep, should be working. Here comes the bullet. You ready? I'm ready. Three, two, one, zero. (bullet whacks metal) What do we see? 5.8 milliseconds. 5.8. Is that what you see? Yeah? 5.8 milliseconds. 5.8 plus or minus 0.1. So out comes the average. Call it speed
or call it velocity it's the same thing
in this case. 148.5, 5.8, and I have to
convert it to meters per second. That brings it at 256,
plus or minus. Now you come in here,
with your plus or minuses. This is a one point...
one-third percent error. It's negligible to this one. One out of 58 is about 1.7% so this is the only one
we have to worry about so the uncertainty in there
is about 1.7%. It's less than two-- that's what I wanted
and it gives me an error of about four meters per second. And so this is the result. And you see,
it's only meaningful because we have a good idea
about the uncertainties in the measurement. Just as we introduced
average velocity now I am going to introduce
average acceleration. Notice that the velocity changes
here throughout time. And that brings me
to the next part the logical part, namely,
that we are going to introduce an average acceleration and with a little bit
of imagination you can probably guess
what that looks like. The average acceleration
between time t1 and time t2 would then be the velocity
at time t2 minus the velocity at time t1,
divided by t2 minus t1. And the dimension is lengths
per seconds per time squared so it's meters
per second squared. This is done for
a one-dimensional situation. This number can be larger than
zero, it can be equal to zero and it can be smaller than zero. In our case, t1 to t2 here notice the velocity is zero
as a start. And it begins to increase because this angle of alpha
increases. It's the angle that matters. The angle increases,
so in our case from t1 to t2 the average acceleration
is larger than zero. Look at the angle. However, if you take the average
acceleration between t1 and t5 that is smaller than zero because here
the velocity is zero but here
the velocity is negative. So if you substitute
that in there you get an average acceleration
which is smaller than zero. So the signs in the velocity and the signs in average
acceleration depend crucially on how I have defined
my increasing value of x not where I choose
my zero points. If I reverse the direction
of increasing x then all my signs will change. So you can also write down then that average acceleration,
if you like that is delta v divided by delta t but you must be careful because
the delta v is sign-sensitive. You must obey
your sign convention. I have here a tennis ball and I can bounce this tennis
ball, I can throw it down. And let us assume,
just for simplicity that it hits the floor
at about five meters per second and that it's a very,
very good tennis ball and that it also bounces back with a speed of
about five meters per second. I will choose this to be
my increasing value of x and so it hits the floor
like this. That means the velocity
at which it hits the floor is minus five meters per second. It bounces off, there it comes and it goes up with
plus five meters per second. I call this v1
and I call this v2. So what, now, is
the average acceleration? Well, I would have to know
the time that it takes for this change in direction. In other words, we call that
the impact time. I would say, in this case,
the impact time delta t is probably about
a hundredth of a second and so my average acceleration
would be v2 minus v1-- that is
plus five minus minus five-- that is ten divided
by ten to the minus two and that is plus 1,000 meters
per second squared. I have observed carefully
the signs. If now I say,
"Aha, I don't like this I want to go this-- the value
of increasing x." No big deal. This will become a plus,
this will become a minus and then this would
become a minus. So then the acceleration is minus 1,000 meters
per second squared. I have also here a tomato
and I have here some eggs. Now, imagine now
that I throw the tomato down or, for that matter, the egg and that they hit the floor
at five meters per second. I could do that. They would not come back up. They would go...
(blows raspberry) So therefore the change
in velocity would not be ten-- apart from the sign
that you have to think about-- but it would only be
five meters per second. The impact time
would probably be much longer maybe a quarter of a second. So therefore the average
acceleration during the impact would then be only five
divided by one quarter... would be something like
20 meters per second squared. Now, whether you call it plus or whether you call it minus
20 meters per second squared depends on your convention
of what you call increasing x. But the eggs and the tomatoes
don't care what you call minus
and what you call plus. Whether the acceleration is minus 20 meters
per second squared or plus 20 meters
per second squared you'd better believe it,
the egg will break. So it's only in your convention
that it matters but, of course,
the physics will not change. The eggs couldn't care less what you have chosen
for your sign convention. Something breaks because the magnitude of
acceleration becomes too high. That's why something breaks. A few days ago, I saw
a Sherlock Holmes movie and there was a guy
who fell on the floor-- marble floor-- hit his head,
was lying there motionless. And here was Watson, and
Watson said to Sherlock Holmes "What happened?" Sherlock Holmes
walks over to the guy touches him and he says,
"He crushed his skull." He looked very intelligent,
I must say, when he said that. "He crushed his skull." And I said, "Gee, that's
really physics in action-- It's 801 all the way." (class laughs ) A modest... a really modest
velocity when he hits the floor but he hit the floor
like a billiard ball. The guy was bald, for one thing and so the impact time
was very short. And when the impact time
is short even if you hit the floor
with a modest speed the acceleration is high...
(blows raspberry ) And that was too much and so that's
why his skull was crushed. So what matters is this changing
velocity and the impact time. We now want to make one last
step from average acceleration. We want to go
to the acceleration at any moment in time just the way we did that
with velocity. And that now is a natural step. The acceleration
at any moment in time will be the limit for delta t
goes to zero for v measured at t plus delta t
minus vt divided by delta t. That is
the instantaneous acceleration. And this, you will recognize is the first derivative
of velocity versus time which is also
the second derivative of position versus time. And so here comes
the second equation that I really want
you to remember forever and ever and ever that the acceleration
is dv/dt which is also d2x/dt squared. We can go to our plot and we can
ask ourselves the question now: where is the acceleration zero,
where is it larger than zero and where is it
smaller than zero? Because this value can be
larger than zero, equal to zero and smaller than zero. And now you have
to be very careful when you try to derive that
from this plot. You have to be very careful because you and I have no good
feeling for second derivatives. Velocity is easy-- all you have to do
is looking at alpha. But when it comes
to the second derivative you have to see
how alpha is changing. Well, right here,
the velocity is not changing so the acceleration
everywhere here must be zero. Here the velocity is increasing so the acceleration must be
larger than zero here. Here, the velocity
is almost constant-- it's almost a straight line. What does that mean
for the acceleration? Zero, exactly. Here, when it makes
this rounding curve the velocity is positive here,
but on this side it's negative so what does that mean
for the acceleration? Negative, you got it. And so you can now roughly find where the acceleration
is positive where it's negative
and where it is zero. Let's do
a straightforward example the way that you could expect
it on an assignment or, if you were
extraordinarily lucky you might even get something
like that on an exam. Very straightforward. I'm going to give you the
position x as a function of time and then ask you lots
of questions about it. So this example
is a working example-- x equals eight minus 60
plus t-squared. So this tells you where the
object is at any moment in time and let this be in meters. What now is the velocity
at any moment in time? Well, that's
the derivative dx/dt and I use the following--
x equals t to the power n. Then, as most
of you should know dx/dt is then n times t
to the power n minus one. That's all I'm using. So the derivative of eight
is zero. I get here minus six,
I get here plus 2t-- this would be
in meters per second--- and the acceleration... I have to take the derivative
of the velocity, I get plus two. So notice that the acceleration is constant in time,
is not changing but the velocity is changing. Well, at time t equals zero... just, I will start
to probe a little bit. I want to get a feeling
for what this object is doing. At time t equals zero,
x is plus eight The velocity is minus six
meters per second and the acceleration
equals plus two. I can also ask myself
at what time does x = 0? What are the times
that x is zero? Well, I have to solve
this second-order equation which is something that
you've all done in high school and you will find that
that's the case when the time is plus two
and when the time is plus four. Take the plus two...
that makes this four. 4 + 8 = 12, minus 6 x 2,
that's 0. So you see the 2 works and
you check that the 4 also works. Just for my curiosity,
when is the velocity zero? Oh, that's easy-- that's when this equation
is zero so that's when t equals three. What is, at that moment,
the position? Oh, I substitute
t equals three in here and that gives me minus one. x = -1. So now I'm ready to plot x
as a function of t. It's, of course, a parabola and I use this information
that we have just derived. So here comes my plot. Let this be
increasing value of x let this be eight
and let this be minus one. This is the time axis. I have a zero here and so I want to cover,
let's say, about six seconds so I have 1, 2, 3, 4, 5, 6. Now I am going to use
this information in order to give you a curve
which is similar to that one except this is a simple one--
this is just a parabola. So I know
that at time t equals zero the object is at position eight. I know that x is zero...
that x is zero at the time 2 and at the time 4 so the object is here
at this time and at this time. And I know that
at time t equals three it is at position minus one. And I also know
that the velocity is zero so we can check that. And so if I make this plot now then we would get a curve
that's sort of like this and yes, indeed, notice,
the velocity here is zero. The angle alpha equals zero. The object starts out
at t equals zero with a negative velocity. You can see that-- the object
at t equals zero is here. This is where the object is,
I hope you realize that. The object is never here. This is the road, this is
the one-dimensional track on which the object is sitting. The object is here and it starts
going in this direction. If it starts going
in this direction the velocity must be
less than zero and indeed it is,
it's minus six. But there is the acceleration which is plus two
in this direction. The acceleration says,
"I don't want you to go down. I want you to go up!" Well, the velocity says "Sorry, all I can do
is slowly, slowly change" and that's what it's doing. It's slowly changing
the velocity and there comes a time
that the velocity is zero so the object goes down,
the velocity changes and when it is
at position minus one it has come to a grinding halt
and now it is returning. This positive value of a
is now increasing the velocity and that's what you see. I therefore bet you a nickel that if you substitute,
in that equation, t equals four that the velocity better
be positive. It has changed
from a minus sign to a plus sign because of
this positive acceleration. I bet you a nickel
t equals four. What is x... uh, what is v? We want to know v. 8 - 6 + 2 meters per second. You see? Physics works-- v is now
plus two meters per second. So all that information
is in there but I want you to be able
to also digest it. Don't look at that curve as just some dumb parabola,
some dumb curve. Try to imagine
what is happening and only then
do you get some insight. Then you really begin
to get it in your brains. I now would like to write down,
in most general form the equation for the position
and the velocity as a function of time
for a one-dimensional motion whereby the acceleration
is constant. So it's going to be
one-dimensional again and we have a is going
to be a constant. And so the equation
that I write down is the most general way
that I can write it down. So we're going to get
x equals some number C1 plus some C2 times t,
plus some C3 times t squared. And notice... oh,
I already erased my example. My example is gone but you would have seen
this was an eight before and here we had...
uh, what did we have? Minus... we had minus 6t
and we had plus 1t squared. So you recognize these three...
I can now take the derivative and so I get C2 plus 2C3
times t and then I get
the acceleration equals 2C3. And now we get some insight
into these quantities. Clearly, x1... C1
is the position of x at time t equals zero for which we often write
an x zero. Because when t is zero,
that is where x is. C2 is really the velocity
at time t equals zero because when t is zero,
that's when C2 is v. And the acceleration is
now changing with time. It's 2C3, therefore C3
is half the acceleration. So this gives you some insight in the meaning
of these quantities and you can see... you can read
now, some physics in there. C1, C2, and C3
can independently be either zero, or larger
than zero, or negative. It makes no difference--
each one of these combinations is a valid possibility
in physics. When we have gravity an object is influenced by
the gravitational acceleration and the gravitational
acceleration is a constant. And we write, often for that gravitational
acceleration, the letter "g". Whether I drop an object
or throw it vertically up or I throw it vertically down,
it's all one-dimensional. It becomes two-dimensional
when I throw it at an angle. I keep it one-dimensional the acceleration is
always the same and that g--
gravitational acceleration-- in Boston is 9.80 meters
per second squared and it varies a little bit
for different places on Earth. This gravitational acceleration
is independent of the mass
of the object that I drop of the speed of the object of the chemical
composition of the object of the size of the object
and of the shape of the object assuming that we have
no air drag assuming that these experiments
are done in... in vacuum. Is it obvious that
the gravitational acceleration is independent
of all these quantities? By no means. Is it true? We think so, but I want you
to appreciate that it is not obvious and it can not be proven
from first principles. Remember, last time we dropped
an apple from three meters and we dropped another one
from one and a half meters. And in your second assignment,
which you haven't seen yet I'm asking you to calculate the gravitational
acceleration for me using these both experiments. And, of course,
I want you to also tell me what the uncertainty is
in your final answer. And I'd like to help you
a little bit to set it up and also to get these equations
in terms of gravity. Whenever we deal with gravity,
we get the g in there. So suppose here is the object
at time t equals zero. It was the apple, and I call
that position x zero. I call that zero, I'm free
to choose my zero position and I drop it zero speed. I just let it go, because that's
the way we did it in class. The object goes down
and it hits the floor. Well, the general equations,
now, which deal in gravity... If I call this
the increasing value of x... You can choose it differently. This is my choice today...
is the following. x equals x zero plus v zero t
plus one-half g t squared and g now is 9.80 meters
per second squared. The velocity,
at any moment in time equals v zero plus gt and the acceleration
is constant-- it's simply g. Now, in my case, I have chosen
t equals zero, x zero, zero and I have chosen this zero,
so these go. And so you see
that when the object is here-- which is three meters
below this point-- and you know the time,
how long it took to get there that you can now calculate "g" because x would be then
three meters. That's when it's here. We made a measurement in class how long it took,
so you know the time and so you can come up
with a value for g. And you can do that
for both measurements and, of course, I want you
to tell me, also what the uncertainty is
in those measurements. Remember that we derived,
last time, that C... that the time that it takes
for the apple to fall was C times the square root
of h over g and we never knew
what that C was. I did a demonstration
to show you that the time is proportional
to the square root of h. We never knew what that C was. Now you know, because now you
have the equations here and you see that that C simply
was the square root of two. But I could not derive that
from my dimensional analysis. Now I want you to relax
and, at the same time get a little bit alert
for a change. Look at this situation,
v equals gt. That means
when I drop an apple-- and I'm going to drop
another one today-- that the velocity
increases with time. So if I strobe this apple
while it was falling I would see the separation,
when it strobes to increase with time, because
the velocity goes up with time. I have here an apple, or
I am going to put an apple up about three meters from
the floor-- three meters. So the height is three meters,
approximately. We know from last time,
remember, we did it it was about 780 milliseconds
to hit the floor. I will just round it off
and I think about it... about eight-tenths of a second,
just to get an idea. If I flash it, if I strobe it
twice per second-- we call that two hertz-- so my strobe is
two times per second. Then I should hit that ball,
when it's falling twice with my strobe light. I don't know
where it is, though because when we strobe it
and when I let the apple go the two are not synchronized,
so maybe the first time that the light blinks,
it may be here and the second time,
it may be here. But it's also possible
that the first time it's here and the second time, it's there. And so the first thing
I want to do is to test your alertness. We will blink. You will tell me
where you see them. But we will take a picture. We will take a picture
which will show us exactly where
those two balls were. So that's
the first alertness test. So get ready for this,
and then we will do a second one which is even more intriguing. So now I have
to first lower this velvet so that we get
a nice dark background. There we go. (whooshes ) Wow, with my fingerprints on it,
it's not so black any more. There it is...
that's the background. Oh, what am I doing? I need the ladder again--
I have to bring the apple up! Friday's always
a bad day for me. Okay... so now I am going
to bring the apple up. There's some metal here,
there are electromagnets and so I throw a switch here so that the electromagnet
is activated. Very similar
to what we did last time. We have to put the apple up
and the apple is hanging there. There we go. So now I have to start
the, uh... The strobe. That's about two hertz, that's
about two flashes per second and I'm going to make it
pitch black. Pitch black. All the lights go off. I will count down 3, 2, 1, 0 and Bob, there,
who is behind the camera will open the shutter
when I say "one." And when I say "zero",
the ball will fall. So you may only see the ball
in its highest position. That may not count there,
of course because it makes two flashes
in the time that the shutter is open
and that I drop it. Okay, if you're ready,
I'm ready. Make it as dark as we can. Bob, are you ready? Class ready? CLASS:
Yes. LEWIN:
Everyone ready? You don't look ready. LEWIN:
Okay... three, two, one. That was zero. So let's look at this again
in slow motion. So now we are developing
the picture and I would like you to tell me
where you saw the balls. Where were they, roughly? Where was the first one? How much... how much below
the highest point? Only this much? The first one. And then the second one
was pretty low, then. (class murmurs ) Okay, sounds interesting. We'll take a look. While the picture is developing I'm now going to test
your real alertness. I'm going to strobe it
with an unknown frequency... unknown to you. I will tell you a secret--
it's a higher frequency. You're going to see
more balls on the way down. I'm not going to ask you
where they are, exactly. All I want you to tell me,
afterwards, how many you saw. That's all. So count them as it falls. You know we have
only 0.8 seconds to count. Bob, how did
the picture come out? Wow, you're good! Whoa, you're good. It was very high, actually... the first... the first flash,
very high. You see, it's...
you did very well. We're going to start, now,
with the second part. Is the audio restored? Should be. So, I activated
the magnet again. There it is. Oh, goodness! Working? Okay, thank you, Bob. Okay, Bob,
if you're ready, I'm ready. We're going to make it
as dark as we can. So all I want you to tell me,
how many balls will you see? Alright, ready? Bob, you're ok? Three, two, one... (class laughing ) Well? Who saw three? Four? Four, I want to know four. Five? Five, here's a five,
there's a five. Another five? Who saw six? Wow... seven? Eight? Nine? Ten? Eleven? Who just saw a blur? (class laughs ) Those are the real winners,
I think. Well, I'll tell you,
it was ten hertz. Since it was 0.8 seconds,
depending upon where you hit it how lucky you are,
I will show you. You will either see seven
or maybe eight balls but it was a good test. And for those of you
who thought that it was only... that only saw five, there you
see them, let's count them. Let's count them together. One, this is one. Two, three, four, five, six,
seven, this is a bounce. So for those who saw five,
I would say "Take some rest this weekend,
you need it" and I'll need it, too. See you Monday.
