Question Video: Using Resultant Motion to Find an Average Speed Physics • First Year of Secondary School

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 runner’s 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 field’s edges. What was the second runner’s 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 we’ll address those parts one at a time. But first, let’s begin by clearing some space and summarizing the information that we’re given in the question.

We’ve got two different runners running in a field, so let’s refer to the first one mentioned in the question as runner one and the second one as runner two. We’re 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 that’s 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 we’ve drawn shows the route that he runs. We’re told that runner one has an average speed of eight meters per second. We’ll label this speed as 𝑣 subscript one, where the subscript one is used to indicate that this quantity is referring to runner one.

The other information that we’re given is about the second runner, who we’ve labeled as runner two. We’re 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 she’s 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. Let’s label this time as 𝑡. While we’re labeling things, let’s also give a name to the distance covered by each runner. We’ll call the distance run by runner one 𝑑 one and the distance run by runner two 𝑑 two. The question doesn’t give us the values of 𝑡, 𝑑 one, or 𝑑 two. We’ll be able to work these things out. But before we get onto that, now that we’ve summarized all the information that the question gives us, let’s remind ourselves what the first part of the question is asking us.

Okay, so this first part of the question says “What was the second runner’s average speed? Answer to two decimal places.”

We can recall that if an object moves a total distance of 𝑑 in a total time of 𝑡, then the average speed 𝑣 of that object is equal to 𝑑 divided by 𝑡. 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, we’ll be able to work out the value for the total distance that runner two covers. However, at the moment, we don’t 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 𝑡.

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 𝑡 to complete their respective routes. Since we know runner one’s average speed and we can use their route to calculate the total distance that they cover, then we’ll be able to use those values along with this equation in order to calculate the value of the time 𝑡. Then this value of 𝑡 will be the same for runner two, and we can calculate the distance that she covers from her route. And at that point, we’ll be able to use these values of distance and time along with this equation to calculate runner two’s average speed.

So let’s make a start on this process by calculating the distance 𝑑 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 𝑑 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 𝑑 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 𝑡 that they take. To do this, though, we’ll need to rearrange this equation to make 𝑡 the subject. The first step is to multiply both sides of the equation by 𝑡 so that the 𝑡’s in the numerator and denominator on the right-hand side cancel each other out. Then we divide both sides of the equation by 𝑣 so that on the left-hand side the 𝑣’s cancel out. This leaves us with an equation that says time 𝑡 is equal to distance 𝑑 divided by speed 𝑣.

Applying this equation to runner one, we can simply add a subscript one to the quantities 𝑑 and 𝑣. And then we can substitute in our values for 𝑑 one and 𝑣 one in order to calculate the time 𝑡. When we do this, we get that 𝑡 is equal to 120 meters divided by eight meters per second. This works out as 15 seconds. So we now know the time 𝑡 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 𝑑 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 Pythagoras’s theorem.

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

From Pythagoras’s theorem, we know that ℎ 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 ℎ 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 𝑑 two traveled by runner two is three times the length ℎ of this hypotenuse. So 𝑑 two is equal to three times ℎ. And substituting in the value that we calculated for ℎ, we find that 𝑑 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. Let’s clear some space so that we can do this. Let’s label the average speed of runner two as 𝑣 two. And we know that this is equal to the distance 𝑑 two divided by the time 𝑡. If we substitute in our values for 𝑑 two and 𝑡, we get that 𝑣 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 runner’s average speed was 8.94 meters per second.

Now let’s 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 we’re told to think about the second runner running between two opposite corners of the field. So we’ll 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 that’s 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 runner’s 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 that’s the component represented by this arrow here, which we’ve labeled as 𝑣 subscript 𝑒𝑤, where the 𝑒𝑤 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 we’ll label as 𝜃, must be the same as the corresponding angle in the triangle that we’ve 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 𝜃. We’ll consider a general right-angled triangle with one angle of 𝜃, a side adjacent to this angle of length 𝑎, a side opposite this angle of length 𝑜, and a hypotenuse of length ℎ.

For this general triangle, we have that sin 𝜃 is equal to 𝑜 divided by ℎ, cos 𝜃 is equal to 𝑎 divided by ℎ, and tan 𝜃 is equal to 𝑜 divided by 𝑎. In the case of the triangle from the field, the side adjacent to the angle 𝜃 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 𝑎 and 20 meters as our value of 𝑜, and then use this equation to calculate the value of the angle 𝜃.

First, though, we’re going to want to make 𝜃 the subject of the equation. To do that, we take the inverse tan of both sides of the equation. The inverse tan of tan 𝜃 is simply 𝜃. And so we have that 𝜃 is equal to the inverse tan of 𝑜 divided by 𝑎. If we then substitute in our values of 20 meters and 40 meters for 𝑜 and 𝑎, respectively, we have that 𝜃 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 𝜃 is equal to 26.565 degrees.

We know that this angle 𝜃 is the same as the angle in our triangle of velocities. And so we can use it to calculate the east-west velocity component 𝑣 subscript 𝑒𝑤. In this velocity vector triangle, we can also identify the side adjacent to the angle 𝜃, so that’s our 𝑎, the side opposite this angle, so that’s 𝑜, and the hypotenuse, ℎ. In this case, the value of ℎ, the length of the hypotenuse, is the magnitude of the velocity, so that’s 8.94 meters per second. The value of 𝜃 is our value that we calculated of 26.565 degrees. And we’re trying to find the length of the side of this triangle that’s adjacent to the angle 𝜃. So that’s our side 𝑎, which is the component of the velocity that’s parallel to the east-west side of the field.

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

Now, in this case, 𝑎 is our value 𝑣 subscript 𝑒𝑤, and we can go ahead and substitute in our values for the hypotenuse ℎ and the angle 𝜃. Then when we evaluate the expression that we get for 𝑣 subscript 𝑒𝑤, we get a result of 7.996 meters per second. We’re told to give our answer to two decimal places. And so, to this level of precision, we have that the second runner’s average speed parallel to the east-west running sides of the field is 8.00 meters per second.

Okay, let’s 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 we’ve done most of the work already. We’re being asked to find the second runner’s 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 𝜃, 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 𝜃. We’re trying to find 𝑜, the length of the opposite side. That means that we’re going to want to use this equation here, which relates the quantities 𝜃, 𝑜, and ℎ. If we take that equation and multiply both sides of it by the hypotenuse ℎ, then the ℎ’s on the right-hand side cancel. And we find that 𝑜 is equal to ℎ times sin 𝜃. In this case 𝑜 is 𝑣 subscript 𝑛𝑠, the velocity component along the north-south running side of the field.

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