WEBVTT
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This is College Physics
Answers with Shaun Dychko.
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A rock is dropped into a deep well
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and the sound of it hitting the
bottom is heard 2 seconds later
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and our job is to calculate what
the height of the well must be.
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In Part A, we assume that we hear
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the sound at the very moment
when the rock hits the water.
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So we assume that it takes no time
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for the sound to get back
to our ears from the water.
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So that's a bit of a simplification.
It makes the work a lot easier.
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But it works out to making our
calculation for the well depth
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a bit bigger than it should be.
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So let's say that the final
position of the rock
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will equal the initial position plus
its initial velocity times time,
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plus one half times acceleration
times time squared.
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The final position is zero and the
initial position is we don't know.
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That's what we want to find 'cause that
will also be the height of the well.
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The initial velocity is zero
because the rock is just dropped,
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it is not thrown down.
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Then we have this plus one half *at*
squared. The term is still there.
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So we'll subtract one half
*at* squared from both sides
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and switch the sides
around leaving us with
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*y naught* equals negative
one half *at* squared.
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So it's negative one half
times the acceleration
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due to gravity of negative 9.8
meters per second squared,
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times the time of
2.0000 seconds squared,
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giving us 19.6 meters is
the depth of the well
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assuming we hear the sound instantly
when the rock hits the bottom.
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Now, in part B, we make it a
little bit more sophisticated
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our analysis
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because we're going to say
that it does take time
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for the sound to go from the
water back to our ears.
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So this 2.0000 seconds
time that we measure
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is going to be the total
of two time intervals.
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One time interval *t one
we'll call it **delta t one*,
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delta meaning interval
or difference.
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This *delta t one* is going to be
the time it takes for the rock
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to hit the water and we don't
know what that time is.
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Then *delta t two*
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is going to be the time it takes for sound
to go from the water back to our ears.
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What we know is the total of
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*delta t one** plus **delta t two*
and that is 2.0000 seconds.
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Anyway, so we write down
the stuff that we know.
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We know the speed of sound
is 332 meters per second.
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We know acceleration. We know
initial velocity is zero
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and we have our final position is
zero at the bottom of the well
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and our initial position is y
naught which we want to find.
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We also don't know what *delta
t one** is, nor **delta t two*.
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Okay. So we write
down some equations
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expressing relationships
between different quantities
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and then we combine equations
together to solve for the unknowns.
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So we don't know *t one*,
we don't know *t two*,
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and we don't know *y naught*. So
because there are three unknowns
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we are going to need three equations.
That always works that way,
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that you need as many equations
as there are unknowns
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to solve the system of equations.
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It's called a system of equations
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when you have several different
equations relating to the same problem.
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Okay. So one equation we can use
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is that this *y naught*
equals the speed of sound
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multiplied by the time it takes
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to get from the water
to the person's ear.
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There is no acceleration involved because
this is sound we're talking about.
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It is not affected by gravity.
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So we have *y naught* equals
*vs* times *delta t two*.
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Then we know that the sum
of the two time intervals
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is 2 seconds
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and now we'll make an expression
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for this time interval
when the rock is dropped
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and it's accelerating down
to the bottom of the well.
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We'll say that the final position *y*
equals initial position *y naught*
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plus *v naught* times *delta t one*,
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plus one half times acceleration
times *delta t one* squared.
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So this is *delta t one*
here which we don't know,
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the interval of time from when the rock
has dropped until it hits the water.
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Now we'll simplify this a bit by getting
rid of the terms that are just zero.
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*v naught* is zero so this term disappears
and the final position is zero
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so we have zero equals *y naught* plus
one half *a delta t one* squared.
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Now we'll do some substitution
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to get rid of one of
our variables here.
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This is an equation with two
unknowns and so we can't solve it.
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We don't know what *y naught* is and
we also don't know *delta t one*.
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So there's nothing we can do unless
we have other equations to help us.
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So we want to make a
substitution for *delta t one*
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and if we can make an expression
that is in terms of *y naught*,
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and that being the only unknown
in our substitution here,
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then we will be able
to solve this equation
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because it will contain only
one unknown, being *y naught*.
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So let's work on that. This
equation is going to be helpful.
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It's going to say *delta t one*
is two minus * delta t two*
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but that's not quite good enough because
we also don't know *delta t two*.
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But we can use equation one to get
an expression for *delta t two*
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in terms of *y naught*
which is good.
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So it's *y naught* divided by *vs*
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where we divided both
sides by *vs* here.
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Then looking at equation two
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we substitute for *delta t two* and
replace it with *y naught* over *vs*.
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So we have *delta t one* plus *y
naught* over *vs* equals two.
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Now we have, after you subtract *y
naught* over *vs* from both sides,
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we have *delta t one* equals
two minus *y naught* over *vs*.
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This is good stuff because if
we plug it into equation three,
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we will have an equation
containing only one unknown.
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So that's what we do
in this line here
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where I've re-written equation
three, called it three b
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and I've made a substitution
for *delta t one*
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as two minus *y naught*
over *vs* instead.
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The only thing we don't
know in this equation
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is *y naught* and so
we can solve it.
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We know what the
speed of sound is
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and we know what the
acceleration due to gravity is.
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Okay.
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I'm going to square this binomial
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and here's the pattern for
squaring a binomial by the way.
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If you have one term minus
another term all squared,
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the answer to that is going
to be the first term squared,
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and minus two times the first
term, times the second term,
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and then plus the
second term squared.
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So we have the first term being
two, so two squared is four,
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minus two times the first term,
so two times two makes four,
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times the second term
*y naught** over **vs*,
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and then plus the second
term squared, *y naught*
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squared over *vs*squared.
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Then we multiply through by the one
half *a* into each term in the brackets.
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We have one half times
four makes *two a*
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and then minus two a
* y naught* over *vs*
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and then plus a *y naught*
squared over two *vs* squared.
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We're going to collect the
linear terms together
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because we want to make this into
the form of a quadratic equation
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because we have a quadratic formula
that we can use to solve it.
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So we have the
squared terms first.
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We have a *y naught* squared
over two *vs* squared.
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I just moved it there,
nothing special about that.
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The next part involves
a little bit of work
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where I collected these
two terms together
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by factoring out the
common factor *y naught*.
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That gives us bracket
one minus two a over *vs*
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then we have plus two a -- this term
is just moved there -- equals zero.
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Then substituting in numbers,
we have negative 9.8
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*y naught* squared over two times the
speed of sound, 332 squared --
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normally I'd put units into everything
but there is so much numbers,
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so many numbers here that
that's too much work.
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So I'm just going to write
numbers in this case.
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Most of the time
I'll still do units.
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Then bracket one minus two times
negative 9.8 over 332 *y naught**,
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plus two times negative
9.8, all equals zero.
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Then you do calculations there and
it works out to these numbers
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which we then substitute
into the quadratic formula.
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So we have the coefficient
of the linear term negative,
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plus or minus the square root of the
coefficient of linear terms squared,
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minus four times the coefficient
of the quadratic term,
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multiplied by the constant term,
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and divided by two times
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that coefficient of **y naught*
squared, negative by the way.
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This all works out
to 18.5 meters
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and it turns out that you
need to use the plus here.
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You can't really know that until
you try one and then the other.
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If you use a minus, you get a
really big negative number
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which doesn't have a
physical interpretation here
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and so we take the
positive answer.
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So there's the height or there's the
depth of the well, 18.5 meters.