WEBVTT
00:00:03.320 --> 00:00:05.420
You're standing at a
traffic intersection.
00:00:05.420 --> 00:00:08.119
And you start to accelerate
when the light turns green.
00:00:08.119 --> 00:00:12.490
Suppose that your acceleration
as a function of time
00:00:12.490 --> 00:00:18.200
is a constant for some time
interval t less than t one.
00:00:18.200 --> 00:00:24.470
And after that, it's zero for
a time after t one less than t
00:00:24.470 --> 00:00:27.340
less than some time t two.
00:00:27.340 --> 00:00:30.290
At the exact same instant
the light turns green,
00:00:30.290 --> 00:00:33.090
a bicyclist is coming
through the intersection.
00:00:33.090 --> 00:00:36.610
And the bicyclist has
some initial speed
00:00:36.610 --> 00:00:40.900
and is braking with an
acceleration of minus
00:00:40.900 --> 00:00:46.980
b two for the entire
time interval t two.
00:00:46.980 --> 00:00:52.650
And at time t two,
the bicyclist comes
00:00:52.650 --> 00:00:58.980
to rest exactly where
you are located.
00:01:01.950 --> 00:01:06.100
And we also know some
initial conditions.
00:01:06.100 --> 00:01:09.830
So our initial conditions
in this problem
00:01:09.830 --> 00:01:13.930
are that you're accelerating
b one at a rate two
00:01:13.930 --> 00:01:15.820
meters per second squared.
00:01:15.820 --> 00:01:19.430
And you do this for time
t one equals one second.
00:01:19.430 --> 00:01:21.490
And the bicyclist comes
into the intersection.
00:01:21.490 --> 00:01:25.382
We'll call that b two naught.
00:01:25.382 --> 00:01:27.090
That's the initial
speed of the bicyclist
00:01:27.090 --> 00:01:29.610
at three meters per second.
00:01:29.610 --> 00:01:35.050
And the question is, what
is the rate of deceleration
00:01:35.050 --> 00:01:38.060
of the bicyclist b two?
00:01:38.060 --> 00:01:41.780
Now this can be quite
a complicated problem.
00:01:41.780 --> 00:01:44.990
So the first thing we want
to do is just make a sketch
00:01:44.990 --> 00:01:47.360
and think about what's involved.
00:01:47.360 --> 00:01:51.020
This problem
involves two objects.
00:01:51.020 --> 00:01:55.350
You and the bicyclist.
00:01:55.350 --> 00:01:59.970
The person-- that's you--
has two stages of motion.
00:02:05.470 --> 00:02:08.580
And the bicyclist only
has one stage of motion.
00:02:12.820 --> 00:02:16.100
So to get started,
it always helps
00:02:16.100 --> 00:02:18.240
to choose a coordinate
system and to make
00:02:18.240 --> 00:02:19.910
some sketches of the problem.
00:02:19.910 --> 00:02:24.220
So let's say we choose a-- it's
all one dimensional motion.
00:02:24.220 --> 00:02:25.260
Two objects.
00:02:25.260 --> 00:02:30.010
One dimensional motion.
00:02:30.010 --> 00:02:34.370
And so we'll pick an origin
at the light at the one
00:02:34.370 --> 00:02:35.850
side of the intersection.
00:02:35.850 --> 00:02:38.850
And we have two objects
which we'll talk about.
00:02:38.850 --> 00:02:41.410
You, x one.
00:02:41.410 --> 00:02:44.860
And the bicycle is x two.
00:02:44.860 --> 00:02:49.880
Actually we don't know yet
who's in front of the other.
00:02:49.880 --> 00:02:52.920
The bicyclist will be
first in front of you.
00:02:52.920 --> 00:02:55.360
So now how do we
sketch the motion
00:02:55.360 --> 00:02:57.630
of these two stages of motion?
00:02:57.630 --> 00:02:59.960
So let's make a sketch.
00:02:59.960 --> 00:03:01.880
And let's start with the person.
00:03:01.880 --> 00:03:04.900
Well, the person-- if we
plotted their position
00:03:04.900 --> 00:03:07.872
as a function of time-- this
would be position in general.
00:03:07.872 --> 00:03:09.330
I'll just draw the
person function.
00:03:09.330 --> 00:03:13.210
They're accelerating
to time t one.
00:03:13.210 --> 00:03:19.500
And then they're moving at a
constant speed at time t two.
00:03:19.500 --> 00:03:22.410
Now the bicyclist is a
little more complicated.
00:03:22.410 --> 00:03:29.130
Because initially the bicyclist
has a-- this at time t,
00:03:29.130 --> 00:03:29.910
person one.
00:03:32.970 --> 00:03:37.930
Initially the bicyclist
has a non-zero slope.
00:03:37.930 --> 00:03:40.790
And they're decelerating.
00:03:40.790 --> 00:03:44.180
And they reach you
with a zero slope.
00:03:44.180 --> 00:03:49.340
So this graph,
this is the x two.
00:03:49.340 --> 00:03:51.340
That is the bicyclist.
00:03:51.340 --> 00:03:53.380
And right here we
have the person.
00:03:53.380 --> 00:03:55.210
x one.
00:03:55.210 --> 00:03:59.840
So now to build a strategy,
we can even look at our graph
00:03:59.840 --> 00:04:02.670
and see that from our
initial conditions,
00:04:02.670 --> 00:04:10.240
we have some special conditions
that the-- our strategy will
00:04:10.240 --> 00:04:14.830
be to-- one-- figure
out what this time is.
00:04:14.830 --> 00:04:19.519
And we know that the
bicyclist at time t two
00:04:19.519 --> 00:04:22.120
has come to a stop.
00:04:22.120 --> 00:04:24.140
So that's one condition.
00:04:24.140 --> 00:04:25.760
And we also know
that the bicyclist
00:04:25.760 --> 00:04:30.270
comes to stop exactly at the
same position as the person.
00:04:30.270 --> 00:04:35.420
x one of t two equals
x two of t two.
00:04:35.420 --> 00:04:37.580
So those are two
conditions that we
00:04:37.580 --> 00:04:41.150
can deduce from all of
this given information.
00:04:41.150 --> 00:04:44.520
And now we comply our
kinematic relationships
00:04:44.520 --> 00:04:47.980
for both the bicyclist
and the person
00:04:47.980 --> 00:04:50.970
and try to see if these
conditions will enable
00:04:50.970 --> 00:04:54.600
us to deduce what b two is.
00:04:54.600 --> 00:04:57.670
So let's begin
with the bicyclist.
00:04:57.670 --> 00:05:01.730
So the velocity of the
bicyclist as a function of time
00:05:01.730 --> 00:05:06.020
is simply the integration
of that bicyclist b t
00:05:06.020 --> 00:05:07.690
prime from zero to t two.
00:05:07.690 --> 00:05:10.220
This is one stage of motion.
00:05:10.220 --> 00:05:11.889
The acceleration is minus b two.
00:05:11.889 --> 00:05:13.680
So this is a very
straightforward interval.
00:05:13.680 --> 00:05:18.520
This is just b of two t two.
00:05:18.520 --> 00:05:19.730
b two.
00:05:19.730 --> 00:05:23.590
This is minus the initial
speed equals that.
00:05:23.590 --> 00:05:28.440
b of t two minus V of
the initial is that.
00:05:28.440 --> 00:05:31.140
And because we want
this to be zero,
00:05:31.140 --> 00:05:33.430
we have the condition
that t two equals
00:05:33.430 --> 00:05:37.330
V two naught divided by b two.
00:05:37.330 --> 00:05:41.350
So that's our first
condition for the bicyclist.
00:05:41.350 --> 00:05:46.460
Now we have to separately solve
for the bicyclist's position.
00:05:46.460 --> 00:05:47.850
That's easy.
00:05:47.850 --> 00:05:54.409
x two of t is the integral of V
two t prime dt prime from zero
00:05:54.409 --> 00:05:55.490
to t two.
00:05:55.490 --> 00:06:00.780
And that's just minus one half.
00:06:00.780 --> 00:06:05.030
We want to make sure that
we get the displacement.
00:06:05.030 --> 00:06:07.320
But x two naught is zero.
00:06:07.320 --> 00:06:16.100
So we have x two at t two equals
the integral of the velocity
00:06:16.100 --> 00:06:22.210
function, which is V
two naught minus b two
00:06:22.210 --> 00:06:26.850
of t prime, d t prime
from zero to t two.
00:06:26.850 --> 00:06:33.820
And so we get b two naught
t two minus one half b two t
00:06:33.820 --> 00:06:35.020
two squared.
00:06:35.020 --> 00:06:38.120
And when we input this
condition in for t two,
00:06:38.120 --> 00:06:45.350
this becomes very simply V two
naught squared over two b two.
00:06:45.350 --> 00:06:49.200
Substituting t two into
each of these expressions
00:06:49.200 --> 00:06:51.300
gives us that relationship.
00:06:51.300 --> 00:06:56.790
So that's the position of
the cyclist at time t two.
00:06:56.790 --> 00:06:58.360
Now this is a
little bit trickier
00:06:58.360 --> 00:07:01.780
to get the position
of the person.
00:07:01.780 --> 00:07:05.960
So in order to do that,
we first find the velocity
00:07:05.960 --> 00:07:07.520
of the person function.
00:07:07.520 --> 00:07:09.210
It's a two stage motion.
00:07:09.210 --> 00:07:12.250
So for the first stage
of motion, the velocity
00:07:12.250 --> 00:07:16.570
two-- the velocity
of person one--
00:07:16.570 --> 00:07:20.110
minus their initial velocity,
which is zero minus one zero.
00:07:20.110 --> 00:07:21.156
That's zero.
00:07:21.156 --> 00:07:27.470
Equals the integral of
b one dt prime from zero
00:07:27.470 --> 00:07:36.430
to t one, which is
just b one t one.
00:07:36.430 --> 00:07:40.450
And this velocity
remains constant
00:07:40.450 --> 00:07:43.240
throughout the next interval.
00:07:43.240 --> 00:07:48.560
So we can write the velocity
function in the following way.
00:07:48.560 --> 00:07:56.810
V one t equals b one t for zero
less than t less than t one.
00:07:56.810 --> 00:08:01.830
And afterwards, a
constant velocity.
00:08:01.830 --> 00:08:06.270
Now this is the function
that we need to integrate
00:08:06.270 --> 00:08:08.170
to get the displacement.
00:08:08.170 --> 00:08:14.810
So let's get ourselves a little
room here and integrate that.
00:08:14.810 --> 00:08:20.600
And we have x of t
is two integrals.
00:08:20.600 --> 00:08:24.110
First from zero to
t one, the velocity
00:08:24.110 --> 00:08:27.100
function during
that time interval.
00:08:27.100 --> 00:08:31.110
And then for the second
time interval dt two,
00:08:31.110 --> 00:08:36.181
the velocity function is
constant b-- this is b one.
00:08:36.181 --> 00:08:40.289
b one t one, dt prime.
00:08:40.289 --> 00:08:42.808
Notice this is not a variable.
00:08:42.808 --> 00:08:45.589
But it is the time at
the end of the interval.
00:08:45.589 --> 00:08:47.130
And when we make
these two intervals,
00:08:47.130 --> 00:08:51.560
we get one half b
one t one squared.
00:08:51.560 --> 00:08:58.640
Let's make this the
velocity at time t.
00:08:58.640 --> 00:09:02.160
This first integral
goes from zero to t one.
00:09:02.160 --> 00:09:04.030
And the second
interval, we're going
00:09:04.030 --> 00:09:07.620
to make this the
position at time t two.
00:09:07.620 --> 00:09:15.910
And we get plus V one times t
one times t two minus t one.
00:09:15.910 --> 00:09:19.880
We have a common term, t
one squared, b, one half b
00:09:19.880 --> 00:09:20.830
one t one squared.
00:09:20.830 --> 00:09:23.780
b one minus b one,
t one squared.
00:09:23.780 --> 00:09:28.820
So this reduces to one
half b one t one squared
00:09:28.820 --> 00:09:32.870
plus b one, t one, t two.
00:09:32.870 --> 00:09:34.610
And that's how we
find the position
00:09:34.610 --> 00:09:38.960
of the person for our interval.
00:09:38.960 --> 00:09:40.630
Let's just review
that to make sure.
00:09:40.630 --> 00:09:43.580
Because we had to get the
velocity function first.
00:09:43.580 --> 00:09:48.040
And then we integrated the
velocity in each time interval
00:09:48.040 --> 00:09:51.070
correctly in order to get
the position function.
00:09:51.070 --> 00:09:53.760
Now we can apply our conditions.
00:09:53.760 --> 00:09:56.740
Notice we already
know t two here.
00:09:56.740 --> 00:09:59.310
And we can now apply
the second condition
00:09:59.310 --> 00:10:05.160
which says that the position
of the bicyclist at time t two,
00:10:05.160 --> 00:10:08.770
which we found to be d
naught squared over two b two
00:10:08.770 --> 00:10:14.090
is equal to the position of
the person at that same time.
00:10:14.090 --> 00:10:19.650
So that's minus one half b one
t one squared plus b one t one.
00:10:19.650 --> 00:10:23.530
Now let's make that
substitution for time t two.
00:10:23.530 --> 00:10:27.520
So that's V two
naught over b two.
00:10:27.520 --> 00:10:33.450
And now our problem is to
solve for this time b two.
00:10:33.450 --> 00:10:35.830
And we're given b one.
00:10:35.830 --> 00:10:37.530
We're given t one.
00:10:37.530 --> 00:10:39.390
We're given V two naught.
00:10:39.390 --> 00:10:42.260
And the only variable
here is b two.
00:10:42.260 --> 00:10:45.240
It's a little bit of
algebra to rearrange terms.
00:10:45.240 --> 00:10:47.910
What I'll do is I'll bring
this term over to here.
00:10:47.910 --> 00:10:51.680
So now we'll just do a
little bit of algebra.
00:10:51.680 --> 00:10:54.410
We have to a b two
we can pull out.
00:10:54.410 --> 00:10:58.810
I have a minus b one
t one b two naught.
00:10:58.810 --> 00:11:03.640
And that's equal to minus
one half b one t one squared.
00:11:03.640 --> 00:11:06.350
And now I can solve for b two.
00:11:06.350 --> 00:11:10.290
And so I get b two
is equal to Vt naught
00:11:10.290 --> 00:11:17.210
squared minus b one t one b two
naught over minus one half b
00:11:17.210 --> 00:11:20.590
one t one squared.
00:11:20.590 --> 00:11:26.680
Now let's just do a
quick dimensional check.
00:11:26.680 --> 00:11:29.750
b times t has the
dimensions of velocity.
00:11:29.750 --> 00:11:31.930
So this is velocity
squared, velocity squared.
00:11:31.930 --> 00:11:33.570
That's OK upstairs.
00:11:33.570 --> 00:11:36.740
b times t squared is
dimensions of position.
00:11:36.740 --> 00:11:39.860
So what we have is meters
squared per second squared
00:11:39.860 --> 00:11:41.360
divided by meters.
00:11:41.360 --> 00:11:43.830
That gives us meters
per second squared.
00:11:43.830 --> 00:11:46.800
So we're pretty confident
that we at least didn't
00:11:46.800 --> 00:11:48.750
make an algebraic mistake.
00:11:48.750 --> 00:11:53.370
And now our last step is to
substitute in the numbers.
00:11:53.370 --> 00:11:56.980
And what we get is, if we
put in the three meters
00:11:56.980 --> 00:12:03.170
a second squared minus b
one times t one times three,
00:12:03.170 --> 00:12:07.710
we get upstairs is minus 3/2.
00:12:07.710 --> 00:12:11.710
And downstairs is
two times one second.
00:12:11.710 --> 00:12:13.090
Two's cancel.
00:12:13.090 --> 00:12:20.380
So we get to 3/2 meters
per second squared
00:12:20.380 --> 00:12:23.680
when we put in the numbers.
00:12:23.680 --> 00:12:25.930
If we wanted to
check our result,
00:12:25.930 --> 00:12:29.456
we can then see
what time we get.
00:12:29.456 --> 00:12:41.400
t two is is three meters per
second divided by our b two
00:12:41.400 --> 00:12:44.570
which is 3/2 meters
per second squared.
00:12:44.570 --> 00:12:47.340
So that's 2 seconds.
00:12:47.340 --> 00:12:51.580
And now the last check would
be to see that the position
00:12:51.580 --> 00:12:53.530
functions correspond to that.
00:12:53.530 --> 00:12:57.440
Let's see if we can just do
that quickly in our heads.
00:12:57.440 --> 00:12:59.570
Our position function
for the person
00:12:59.570 --> 00:13:04.470
is V naught squared
over two b two.
00:13:04.470 --> 00:13:11.110
So that's 9 meters per second.
00:13:11.110 --> 00:13:13.660
9 meters squared seconds
squared over two times
00:13:13.660 --> 00:13:18.000
3/2 meters second squared.
00:13:18.000 --> 00:13:22.590
And that comes
out to x two of t.
00:13:22.590 --> 00:13:25.480
This is a check, is 3 meters.
00:13:25.480 --> 00:13:30.730
And the x one of t.
00:13:30.730 --> 00:13:31.870
We left out the one there.
00:13:31.870 --> 00:13:33.450
We should have had it.
00:13:33.450 --> 00:13:36.090
It's a little more
complicated to put in here.
00:13:36.090 --> 00:13:38.870
But we'll just run the
numbers quickly through.
00:13:38.870 --> 00:13:40.730
Two times one seconds minus.
00:13:40.730 --> 00:13:42.080
That's a minus one.
00:13:42.080 --> 00:13:44.730
Two times one times two.
00:13:44.730 --> 00:13:45.860
That's two times two.
00:13:45.860 --> 00:13:48.390
So this is also three meters.
00:13:48.390 --> 00:13:53.690
And we actually have
the right answer here.