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Question

How many meters are in a marathon race?

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Solution

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A marathon race is a long-distance running event with an official distance of 42.195 kilometers. To convert this distance into meters, we use the basic conversion factor that 1 kilometer is equivalent to 1,000 meters.

Here is the step-by-step solution to calculate the number of meters in a marathon race:
Here is the step-by-step solution to calculate the number of meters in a marathon race:

Step 1: Identify the conversion factor between kilometers and meters.
- 1 kilometer = 1,000 meters

Step 2: Multiply the total number of kilometers in a marathon by the conversion factor.
- Marathon distance in kilometers = 42.195 km
- Conversion factor = 1,000 meters/1 kilometer

Step 3: Perform the multiplication to convert kilometers to meters.
- Number of meters in a marathon = 42.195 km * 1,000 meters/km

Step 4: Calculate the product to find the total meters.
- Number of meters in a marathon = 42,195 meters

Therefore, a marathon race is 42,195 meters long. This standardized distance was established by the International Association of Athletics Federations (IAAF) and has been used in Olympic Games and marathons worldwide since the early 20th century.

Exercise 5-A water trough in the shape of an issosceles triangular prism with width 6 meters and height 3 meters is 9 meters deep

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This is an exercise in related rates.
A trough of water is 9 meters deep and
its ends are in the shape of an isosceles triangle,
whose width is 6 meters and the height is 3 meters.
If water is being pumped in at a constant rate of 6 cubic meters a second,
at what rate is the height of the water changing,
when the water has a height of a 110 centimeters?
They've given us a diagram,
we just need to fill in some of the numbers and variable names.
What I want to do is my first section is usually what I call denote,
that's where I name which variables I'm using and what they mean.
Under the denote here I'll be writing some variable names,
especially those variables which are changing with time.
Let's first of all mark some things in the sketch.
As I recall, this was 9.
I'll mention that the units of length I'm going to take is going to be meters.
There's just 1 place we have to watch out where they've used
centimeters so we have to convert that to meters,
but I'll take length in meters and time in seconds and of course volume in cubic meters.
We've got this as 6 and we've got this as 3 and these are the dimensions of the trough.
Now, the water is changing and the water has a height,
so how about h for height?
It has a width, so how about w for width?
All together, the water has a volume so volume will be in meters cube,
so that will be V and each of these is changing over time.
What I'll write is that V equals volume of the water in
the trough and the units that's going to be in
cubic meters and we're going to have h for the height,
which is the height of the water,
this will be in meters.
Similarly, we'll have w,
which is the width of water and that's also in
meters and then I usually have a section called hence,
cause I derive it from these V prime.
These are all functions of t,
but like I said earlier we're going to skip
the parentheses t just to clean things up and we know that these are functions of time.
V prime is the rate of change of volume and by volume I mean
the volume of the water in the trough and then I have h prime and w prime,
1 is the rate of change of height,
1 is the rate of change of width, I'll just write it in.
There we are.
All these derivatives are with respect to t. The rate of change, this of course,
is the same as these except divided by the unit of time so this
will be in cubic meters a second,
rate of change of height will be in meters per
second and the same with the width in meters per second.
The next section I go for is
what we're given and then after that, what we're looking for.
I've written both of these headings given and looking for.
What we're given is that V prime,
which is the rate of change of the volume.
It's the rate at which the volume is being pumped in,
so V prime equals 6 and that's a constant doesn't depend on t and it's being pumped in.
Yes, positive and we're looking for h prime,
we're looking for the rate of change of h. When I remember it said
a 110 centimeter when h is 1.1 meters.
The next section that we want is an equation which connects all these variables.
The main equation will be the volume of the trough,
which is actually a triangular prism.
It's the area of the triangle times the depth or length.
The volume of this prism will be the area and we're talking about the changing volume,
so it's the blue, the water is the area of this triangle.
Yeah, I'll even shade it,
times the length of it which is what they call depth,
which will be the 9.
Let's write that down and we will get that V is equal to,
select the area of the base times height for any prism now this area of a triangle.
This bit here, this was w,
so area of a triangle is half base times height so we have 1/2 w times h,
but then times 8 which is the unchanging length and they call it depth.
That's our main equation,
but there is another equation I'd like to write.
I just gained experience from the previous question which also
had similar triangles and I'd like to write that as an equation.
What I want to say is that we have this triangle that I've shaded
is similar to this triangle which is the side of the trough.
What I can say is that if I take this width and I divide it by the whole width,
which is the 6 but this is the same by similar triangles to this h,
h in of the water over 3 of the trough for
the numerators of relate to the water and the denominators are from the trough.
I would like to write this as
another equation because otherwise they'll have too many equations here and this will
enable me to write w in terms of h or the other way
around and last thing where your indulgence,
I didn't want to start developing this later.
I can multiply 8 by a 1/2 and let's write this as already as 4 and
now we come to the section on derive where we will get an extra information.
But I look at this and I say,
okay, I want to derive this,
but I really wanted V in terms of
1 variable and preferably h because that's what I'm looking for.
Now, remember now that, that's why I put in
this extra equation from the similar triangles.
Before I derive, just indulge me here,
I'll do a little bit of algebra.
From this, if I multiply both sides by 6 and remembering that 6 over 3 is 2,
I get that w is equal to 2h.
The second thing I could tell now is that if I put w equals 2h here,
I get that V is 2 times 4 is 8.
I get that V is equal to 8,
hh is h squared and this is very useful to me.
Now, I'll do the deriving and I'll derive both of them.
I don't know if I'll need the second 1,
but the first 1 I'll get will be that V prime is equal
to 16h times h prime because it's also a function of t and the other 1,
each w prime is 2h prime.
Next section is to substitute variables that we know.
Let's go for substitute now.
We know everything except h prime,
which is what we're looking for.
We're looking for it when h is 1.1,
so V prime is a constant,
it's positive, it's 6 always,
that's the rate at which the water is being pumped in is equal to 16 is
just a number h is the 1.1 and times h prime.
Now, we just have to take h prime out and we get
h prime is equal to 6 over 16 times 1.1,
is 17.6 and this is equal to I'm just going to look in my calculator.
Approximately 0.341 to 3 decimal places I took it and this is in meters per second.
Finally, I just write something in words, something human-readable.
The height of the water is changing at a rate of approximately
0.341 meters a second and that's more human-readable, I'm just leaving it like that.
We're done, we've answered the question.
This video explains how to solve a related rates problem involving a trough of water. The trough is 9 meters deep, 6 meters wide, and 3 meters high. Water is being pumped in at a constant rate of 6 cubic meters a second, and the goal is to find the rate of change of the height of the water when the height is 1.1 meters. - Variables are denoted: V = volume of water in the trough (m^3), h = height of water (m), w = width of water (m) - Given: V' = 6 (m^3/s) - Looking for: h' (m/s) - Equations: V = 1/2 w h * 9, w = 2h - Derive: V' = 16h h', w' = 2h' - Substitute: h' = 6/16 * 1.1 = 0.341 m/s - Conclusion: The height of the water is changing at a rate of approximately 0.341 meters a second.

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