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 runnerβs 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 fieldβs edges. What was the second runnerβs
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 weβll address those parts one at a time. But first, letβs begin by clearing
some space and summarizing the information that weβre given in the question.
Weβve got two different runners
running in a field, so letβs refer to the first one mentioned in the question as
runner one and the second one as runner two. Weβre 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 thatβs 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 weβve drawn shows the route that he runs. Weβre told that runner one has an
average speed of eight meters per second. Weβll label this speed as π£
subscript one, where the subscript one is used to indicate that this quantity is
referring to runner one.
The other information that weβre
given is about the second runner, who weβve labeled as runner two. Weβre 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 sheβs 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. Letβs label this time as π‘. While weβre labeling things, letβs
also give a name to the distance covered by each runner. Weβll call the distance run by
runner one π one and the distance run by runner two π two. The question doesnβt give us the
values of π‘, π one, or π two. Weβll be able to work these things
out. But before we get onto that, now
that weβve summarized all the information that the question gives us, letβs remind
ourselves what the first part of the question is asking us.
Okay, so this first part of the
question says βWhat was the second runnerβs average speed? Answer to two decimal places.β
We can recall that if an object
moves a total distance of π in a total time of π‘, then the average speed π£ of
that object is equal to π divided by π‘. 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, weβll be able to work out the value for the total
distance that runner two covers. However, at the moment, we donβt
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 π‘.
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 π‘ to complete their respective routes. Since we know runner oneβs average
speed and we can use their route to calculate the total distance that they cover,
then weβll be able to use those values along with this equation in order to
calculate the value of the time π‘. Then this value of π‘ will be the
same for runner two, and we can calculate the distance that she covers from her
route. And at that point, weβll be able to
use these values of distance and time along with this equation to calculate runner
twoβs average speed.
So letβs make a start on this
process by calculating the distance π 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 π 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 π 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
π‘ that they take. To do this, though, weβll need to
rearrange this equation to make π‘ the subject. The first step is to multiply both
sides of the equation by π‘ so that the π‘βs in the numerator and denominator on the
right-hand side cancel each other out. Then we divide both sides of the
equation by π£ so that on the left-hand side the π£βs cancel out. This leaves us with an equation
that says time π‘ is equal to distance π divided by speed π£.
Applying this equation to runner
one, we can simply add a subscript one to the quantities π and π£. And then we can substitute in our
values for π one and π£ one in order to calculate the time π‘. When we do this, we get that π‘ is
equal to 120 meters divided by eight meters per second. This works out as 15 seconds. So we now know the time π‘ 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 π 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 Pythagorasβs theorem.
This says that for a general
right-angled triangle with a hypotenuse of length π and other sides of lengths π
and π that π squared is equal to π squared plus π squared. Or if we take the square root of
both sides, then since the square root of π squared is simply π, then this
equation says that π is equal to the square root of π squared plus π squared. In the triangle that weβve got for
runner two, the side which corresponds to π has a length of 40 meters and the side
which corresponds to π has a length of 20 meters. Letβs label the hypotenuse of this
triangle as β.
From Pythagorasβs theorem, we know
that β 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 β 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 π two
traveled by runner two is three times the length β of this hypotenuse. So π two is equal to three times
β. And substituting in the value that
we calculated for β, we find that π 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. Letβs clear some space so that we
can do this. Letβs label the average speed of
runner two as π£ two. And we know that this is equal to
the distance π two divided by the time π‘. If we substitute in our values for
π two and π‘, we get that π£ 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 runnerβs average speed was 8.94 meters per second.
Now letβs 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 weβre told to think about the
second runner running between two opposite corners of the field. So weβll 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 thatβs 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 runnerβs 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 thatβs the component represented
by this arrow here, which weβve labeled as π£ subscript ππ€, where the ππ€
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 weβll label as π, must be the same as the
corresponding angle in the triangle that weβve 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 π. Weβll consider a general
right-angled triangle with one angle of π, a side adjacent to this angle of length
π, a side opposite this angle of length π, and a hypotenuse of length β.
For this general triangle, we have
that sin π is equal to π divided by β, cos π is equal to π divided by β, and tan
π is equal to π divided by π. In the case of the triangle from
the field, the side adjacent to the angle π 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 π and 20 meters as our value of π, and then use this equation to
calculate the value of the angle π.
First, though, weβre going to want
to make π the subject of the equation. To do that, we take the inverse tan
of both sides of the equation. The inverse tan of tan π is simply
π. And so we have that π is equal to
the inverse tan of π divided by π. If we then substitute in our values
of 20 meters and 40 meters for π and π, respectively, we have that π 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 π is equal to 26.565 degrees.
We know that this angle π is the
same as the angle in our triangle of velocities. And so we can use it to calculate
the east-west velocity component π£ subscript ππ€. In this velocity vector triangle,
we can also identify the side adjacent to the angle π, so thatβs our π, the side
opposite this angle, so thatβs π, and the hypotenuse, β. In this case, the value of β, the
length of the hypotenuse, is the magnitude of the velocity, so thatβs 8.94 meters
per second. The value of π is our value that
we calculated of 26.565 degrees. And weβre trying to find the length
of the side of this triangle thatβs adjacent to the angle π. So thatβs our side π, which is the
component of the velocity thatβs parallel to the east-west side of the field.
Since we know the values of β and
π and we want to find the value of π, then weβre going to want to use this
equation here, which relates the quantities π, π, and β. Letβs clear ourselves some more
space so we can rearrange this equation to make π the subject. To make π the subject, we multiply
both sides of the equation by the hypotenuse β. Then the ββs on the right-hand side
cancel out. And writing the equation the other
way around, we have that π is equal to β times cos π.
Now, in this case, π is our value
π£ subscript ππ€, and we can go ahead and substitute in our values for the
hypotenuse β and the angle π. Then when we evaluate the
expression that we get for π£ subscript ππ€, we get a result of 7.996 meters per
second. Weβre told to give our answer to
two decimal places. And so, to this level of precision,
we have that the second runnerβs average speed parallel to the east-west running
sides of the field is 8.00 meters per second.
Okay, letβs 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 weβve done most of the work already. Weβre being asked to find the
second runnerβs 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 π, 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
π. Weβre trying to find π, the length
of the opposite side. That means that weβre going to want
to use this equation here, which relates the quantities π, π, and β. If we take that equation and
multiply both sides of it by the hypotenuse β, then the ββs on the right-hand side
cancel. And we find that π is equal to β
times sin π. In this case π is π£ subscript
ππ , the velocity component along the north-south running side of the field.
So then substituting in our values
for β and π, we get this expression here. Evaluating the expression gives us
the second runnerβs average speed parallel to the north-south running sides of the
field. And to two decimal places, this is
4.00 meters per second.