### Video Transcript

An athlete runs once completely
around the edge of a rectangular field. The east-west running sides of the
field are 40 meters long, and the north-south running sides are 20 meters long. The runnerโs average speed is eight
meters per second. While the first runner runs, a
second runner runs back and forth between opposite corners of the field and reaches
the opposite corner to her starting point for the second time at the same moment
that the first runner completes his run all around the fieldโs edges. What was the second runnerโs
average speed? Answer to two decimal places. While the second runner runs
between two opposite corners of the field, what is her average speed parallel to the
east-west running sides of the field? Answer to two decimal places. While the second runner runs
between two opposite corners of the field, what is her average speed parallel to the
north-south running sides of the field? Answer to two decimal places.

Okay, so this question has three
different parts to it, and weโll address those parts one at a time. But first, letโs begin by clearing
some space and summarizing the information that weโre given in the question.

Weโve got two different runners
running in a field, so letโs refer to the first one mentioned in the question as
runner one and the second one as runner two. Weโre told that the field is
rectangular with east-west running sides that are 40 meters long and north-south
running sides that are 20 meters long. The first runner, so thatโs runner
one, runs one time completely around the edge of this field. So if we imagine that he starts in
the bottom-left corner and completes his journey in a clockwise direction, then this
orange path that weโve drawn shows the route that he runs. Weโre told that runner one has an
average speed of eight meters per second. Weโll label this speed as ๐ฃ
subscript one, where the subscript one is used to indicate that this quantity is
referring to runner one.

The other information that weโre
given is about the second runner, who weโve labeled as runner two. Weโre told that she runs in the
same field as runner one but that she follows a different path, running back and
forth between opposite corners of the field. So if she starts in this
bottom-left corner, then sheโs running back and forth between here and the top-right
corner. So she runs across the field on
this straight diagonal line. We are told that at the exact same
moment that runner one has completed one circuit of the edges of the field, runner
two reaches the opposite corner to her starting point for the second time.

So the route that runner two covers
during this time is to first go across from the bottom-left corner, her starting
point, to the opposite top-right corner, then back across from the top right to the
bottom left, and finally once more from the bottom left to the top right so that at
the same time runner one has gone all around the four edges of the field and ended
up back at his starting point, runner two has got back to this opposite corner for
her second time. So then the route covered by runner
two is just three times along this diagonal of the field.

We know that each of the runners
takes the same time to run their route. Letโs label this time as ๐ก. While weโre labeling things, letโs
also give a name to the distance covered by each runner. Weโll call the distance run by
runner one ๐ one and the distance run by runner two ๐ two. The question doesnโt give us the
values of ๐ก, ๐ one, or ๐ two. Weโll be able to work these things
out. But before we get onto that, now
that weโve summarized all the information that the question gives us, letโs remind
ourselves what the first part of the question is asking us.

Okay, so this first part of the
question says โWhat was the second runnerโs average speed? Answer to two decimal places.โ

We can recall that if an object
moves a total distance of ๐ in a total time of ๐ก, then the average speed ๐ฃ of
that object is equal to ๐ divided by ๐ก. We want to calculate the value of
this average speed for runner two. From the information given to us in
the question, we know the route that runner two runs along. And this means that in this
equation for the average speed, weโll be able to work out the value for the total
distance that runner two covers. However, at the moment, we donโt
know the value of the time that runner two takes. And so before we can use this
equation to calculate the average speed of runner two, we need to find a way to work
out the value of the time ๐ก.

The important thing to remember
from the information we got from the question is that runner two and runner one both
take the same time ๐ก to complete their respective routes. Since we know runner oneโs average
speed and we can use their route to calculate the total distance that they cover,
then weโll be able to use those values along with this equation in order to
calculate the value of the time ๐ก. Then this value of ๐ก will be the
same for runner two, and we can calculate the distance that she covers from her
route. And at that point, weโll be able to
use these values of distance and time along with this equation to calculate runner
twoโs average speed.

So letโs make a start on this
process by calculating the distance ๐ one that runner one moves. We know that runner one runs once
along each of the four edges of the field. The first edge is 20 meters long,
the second edge is 40 meters, the third edge is 20 meters again, and the fourth, and
final, edge is 40 meters. So the total distance ๐ one that
runner one travels is equal to 20 meters for the first edge plus 40 meters for the
second edge plus another 20 meters for the third edge plus another 40 meters for the
fourth edge. Adding up all these distances, we
find that ๐ one is equal to 120 meters.

Now that we know the total distance
and the average speed of runner one, we can use these values to calculate the time
๐ก that they take. To do this, though, weโll need to
rearrange this equation to make ๐ก the subject. The first step is to multiply both
sides of the equation by ๐ก so that the ๐กโs in the numerator and denominator on the
right-hand side cancel each other out. Then we divide both sides of the
equation by ๐ฃ so that on the left-hand side the ๐ฃโs cancel out. This leaves us with an equation
that says time ๐ก is equal to distance ๐ divided by speed ๐ฃ.

Applying this equation to runner
one, we can simply add a subscript one to the quantities ๐ and ๐ฃ. And then we can substitute in our
values for ๐ one and ๐ฃ one in order to calculate the time ๐ก. When we do this, we get that ๐ก is
equal to 120 meters divided by eight meters per second. This works out as 15 seconds. So we now know the time ๐ก that
runner one takes to complete his route. And we also know that runner two
takes the same time as runner one. So she also takes a time of 15
seconds.

What we need to do now is to
calculate the distance ๐ two that runner two travels. To do this, we can notice that the
diagonal which runner two moves along forms the hypotenuse of a right-angled
triangle. The other two sides of this
triangle are the edges of the field. And we know that the north-south
edge has a length of 20 meters, while the east-west one has a length of 40
meters. To find the length of this
hypotenuse, we can recall Pythagorasโs theorem.

This says that for a general
right-angled triangle with a hypotenuse of length ๐ and other sides of lengths ๐
and ๐ that ๐ squared is equal to ๐ squared plus ๐ squared. Or if we take the square root of
both sides, then since the square root of ๐ squared is simply ๐, then this
equation says that ๐ is equal to the square root of ๐ squared plus ๐ squared. In the triangle that weโve got for
runner two, the side which corresponds to ๐ has a length of 40 meters and the side
which corresponds to ๐ has a length of 20 meters. Letโs label the hypotenuse of this
triangle as โ.

From Pythagorasโs theorem, we know
that โ must be equal to the square root of 40 meters squared plus 20 meters
squared. This is equal to the square root of
2000 meters squared. If we evaluate the square root, we
find that โ is equal to 44.7214 meters, where the ellipses indicate that this result
has further decimal places.

We know that runner two runs along
this diagonal three times, first from their starting point across to the opposite
corner, then back again to the starting point, and finally once more over to the
opposite corner. This means that the distance ๐ two
traveled by runner two is three times the length โ of this hypotenuse. So ๐ two is equal to three times
โ. And substituting in the value that
we calculated for โ, we find that ๐ two is equal to 134.164 meters.

Now that we know the total distance
traveled by runner two and the total time that they take to travel this distance, we
can use those values in this equation to calculate her average speed. Letโs clear some space so that we
can do this. Letโs label the average speed of
runner two as ๐ฃ two. And we know that this is equal to
the distance ๐ two divided by the time ๐ก. If we substitute in our values for
๐ two and ๐ก, we get that ๐ฃ two is equal to 134.164 meters divided by 15
seconds. Evaluating the expression gives a
result of 8.944 meters per second. And since we are asked to give our
answer to two decimal places, then our answer to this first part of the question is
that the second runnerโs average speed was 8.94 meters per second.

Now letโs clear ourselves some
space and look at the second part of the question.

While the second runner runs
between two opposite corners of the field, what is her average speed parallel to the
east-west running sides of the field? Answer to two decimal places.

Okay, so in the first part of the
question, we calculated the average speed of the second runner. Now weโre told to think about the
second runner running between two opposite corners of the field. So weโll consider a part of her
route where she starts at the bottom-left corner and is running along this diagonal
across to the top-right corner. We know that this diagonal forms
the hypotenuse of a right-angled triangle, so we can say that she has a velocity
with a magnitude of 8.94 meters per second, so thatโs the speed that we found in the
first part of the question, and a direction along the diagonal of the field, which
we know is the hypotenuse of our right-angled triangle.

We can think of this velocity
vector as the resultant vector made up of two components. In this diagram, the pink arrow
represents the runnerโs velocity across the diagonal of the field. The two orange arrows represent the
components of this velocity along the east-west direction and the north-south
direction, respectively. This second part of the question is
asking us to find the component of her velocity along the east-west direction. So thatโs the component represented
by this arrow here, which weโve labeled as ๐ฃ subscript ๐๐ค, where the ๐๐ค
signifies that this is the component of the velocity along the east-west side of the
field.

Since we know that the resultant
velocity vector is directed along the diagonal of the field, then the angle in this
triangle of velocity vectors, which weโll label as ๐, must be the same as the
corresponding angle in the triangle that weโve drawn using the dimensions of the
field. In this triangle, we know the
lengths of all three sides. And so we can use it along with
some trigonometric identities to work out the value of the angle ๐. Weโll consider a general
right-angled triangle with one angle of ๐, a side adjacent to this angle of length
๐, a side opposite this angle of length ๐, and a hypotenuse of length โ.

For this general triangle, we have
that sin ๐ is equal to ๐ divided by โ, cos ๐ is equal to ๐ divided by โ, and tan
๐ is equal to ๐ divided by ๐. In the case of the triangle from
the field, the side adjacent to the angle ๐ has a length of 40 meters, and the side
opposite this angle has a length of 20 meters. So we can use 40 meters as our
value of ๐ and 20 meters as our value of ๐, and then use this equation to
calculate the value of the angle ๐.

First, though, weโre going to want
to make ๐ the subject of the equation. To do that, we take the inverse tan
of both sides of the equation. The inverse tan of tan ๐ is simply
๐. And so we have that ๐ is equal to
the inverse tan of ๐ divided by ๐. If we then substitute in our values
of 20 meters and 40 meters for ๐ and ๐, respectively, we have that ๐ is equal to
the inverse tan of 20 divided by 40, which simplifies to the inverse tan of
one-half. And then evaluating the inverse
tan, we find that ๐ is equal to 26.565 degrees.

We know that this angle ๐ is the
same as the angle in our triangle of velocities. And so we can use it to calculate
the east-west velocity component ๐ฃ subscript ๐๐ค. In this velocity vector triangle,
we can also identify the side adjacent to the angle ๐, so thatโs our ๐, the side
opposite this angle, so thatโs ๐, and the hypotenuse, โ. In this case, the value of โ, the
length of the hypotenuse, is the magnitude of the velocity, so thatโs 8.94 meters
per second. The value of ๐ is our value that
we calculated of 26.565 degrees. And weโre trying to find the length
of the side of this triangle thatโs adjacent to the angle ๐. So thatโs our side ๐, which is the
component of the velocity thatโs parallel to the east-west side of the field.

Since we know the values of โ and
๐ and we want to find the value of ๐, then weโre going to want to use this
equation here, which relates the quantities ๐, ๐, and โ. Letโs clear ourselves some more
space so we can rearrange this equation to make ๐ the subject. To make ๐ the subject, we multiply
both sides of the equation by the hypotenuse โ. Then the โโs on the right-hand side
cancel out. And writing the equation the other
way around, we have that ๐ is equal to โ times cos ๐.

Now, in this case, ๐ is our value
๐ฃ subscript ๐๐ค, and we can go ahead and substitute in our values for the
hypotenuse โ and the angle ๐. Then when we evaluate the
expression that we get for ๐ฃ subscript ๐๐ค, we get a result of 7.996 meters per
second. Weโre told to give our answer to
two decimal places. And so, to this level of precision,
we have that the second runnerโs average speed parallel to the east-west running
sides of the field is 8.00 meters per second.

Okay, letโs clear some space one
last time and look at the final part of the question.

While the second runner runs
between two opposite corners of the field, what is her average speed parallel to the
north-south running sides of the field? Answer to two decimal places.

This last bit of the question
should be relatively quick because weโve done most of the work already. Weโre being asked to find the
second runnerโs average speed parallel to the north-south running sides of the
field. So looking at this triangle of
velocities, this time rather than finding the length of the side adjacent to the
angle ๐, which gave us the east-west velocity component, we want to find the length
of the opposite side because this will give us the north-south component.

We know the length of the
hypotenuse, which is the magnitude of the resultant velocity, and we know the angle
๐. Weโre trying to find ๐, the length
of the opposite side. That means that weโre going to want
to use this equation here, which relates the quantities ๐, ๐, and โ. If we take that equation and
multiply both sides of it by the hypotenuse โ, then the โโs on the right-hand side
cancel. And we find that ๐ is equal to โ
times sin ๐. In this case ๐ is ๐ฃ subscript
๐๐ , the velocity component along the north-south running side of the field.

So then substituting in our values
for โ and ๐, we get this expression here. Evaluating the expression gives us
the second runnerโs average speed parallel to the north-south running sides of the
field. And to two decimal places, this is
4.00 meters per second.