Video Transcript
Two children, Sofia and Michael, go running through some fields. They both enter a field that is 36 meters long and 77 meters wide entering at the field’s bottom left-hand corner at the same time as each other. Sophia runs along the edges of the field at an average speed of 5.65 meters per second, first to the top left-hand corner and then to the top right-hand corner. Michael runs straight across the field to the top right-hand corner, but the long grass in the field makes him run at an average speed of only 2.5 meters per second. What is the difference in the time taken for Sophia and Michael to arrive at the top right-hand corner of the field?
Okay, so in this question we have two children running from one corner of a field to another. We are told that the children set off at the same time as each other but each runs at a different speed and along a different route. We are asked to work out the difference in time at which they arrive at the far corner.
Let’s begin by extracting the relevant information from the question and using it to draw a sketch. We are told that the field is 36 meters long and 77 meters wide. We know that the children enter at the bottom left-hand corner of the field. So on our sketch, that’s this point here. Sophia runs to the top left-hand corner and then on to the top right-hand corner. We’ve added Sofia’s route to the diagram in orange. We’re told that Sophia runs at a speed of 5.65 meters per second. We’ll call this speed 𝑆 subscript s, where the subscript s stands for Sofia.
Meanwhile, Michael runs straight across the field to the top right-hand corner. So that’s this diagonal path here. We’re told that Michael runs at a speed of 2.5 meters per second. We’ll call his speed 𝑆 subscript m, where the m stands for Michael.
We’ve now extracted all of the relevant information from the question. Looking at our diagram, we should see that we’ve got a load of information about distances and speeds. And the question is asking us to work out a value of time. So let’s recall that we have a formula which relates the three quantities speed, distance, and time. Specifically, we have that speed 𝑆 is equal to distance, which we’ve labeled 𝑑, divided by time, labeled 𝑡. Now, in this question, we have information about distances and speeds and we are looking to find a value of time. So let’s rearrange this formula to make time 𝑡 the subject.
If we multiply both sides of the equation by 𝑡, then the 𝑡’s in the numerator and the denominator on the right-hand side of the equation cancel each other out. And we have that 𝑆 multiplied by 𝑡 is equal to 𝑑. Then if we divide both sides by 𝑆, then the 𝑆’s in the numerator and denominator on the left-hand side cancel. And so we end up with the equation that time 𝑡 is equal to distance 𝑑 divided by speed 𝑆. This formula means that if we know the distance each child runs and the speed that they run at, then we can calculate the time that each of them takes to arrive at the top right-hand corner of the field starting from the bottom left-hand corner.
Let’s begin with Sophia. We already know the speed. That’s 𝑆 subscript s equals 5.65 meters per second. We just need to work out the distance. We know that the path she takes is along the orange arrows in the diagram. So that’s first to the top left-hand corner and then from there across to the top right-hand corner. So first she runs along an edge that’s 36 meters long, and then she runs along an edge that 77 meters long. So the total distance that she covers, which we’ll call 𝑑 subscript s, is equal to 36 meters plus 77 meters.
Adding together these numbers, we get that the distance 𝑑 subscript s is equal to 113 meters. Now that we know both Sofia’s distance and her speed, we can use this formula here to work out the time that she takes. If we call this time 𝑡 subscript s, then we have that 𝑡 subscript s is equal to the distance 113 meters divided by the speed 5.65 meters per second. Doing this calculation, we get a time 𝑡 subscript s of 20 seconds. So Sofia takes 20 seconds to make it from the bottom left-hand corner to the top right-hand corner.
Now let’s do the same for Michael. As with Sophia, we already know his speed, in this case, 2.5 meters per second. But again, we need to work out the distance that he runs. From our diagram, we can see that Michael runs along the hypotenuse of a right angle triangle. And we know the lengths of the other two sides of this triangle. They are 36 meters and 77 meters. Pythagoras’s theorem tells us that for a general right-angled triangle with a hypotenuse of length 𝑐 and other sides of length 𝑎 and 𝑏, that 𝑐 squared is equal to 𝑎 squared plus 𝑏 squared. Taking the square root of both sides, we have that the length 𝑐 of the hypotenuse is equal to the square root of 𝑎 squared plus 𝑏 squared.
Comparing this general right-angled triangle with the one in the diagram that we drew for the question, then we can see that in our case, 𝑎 is equal to 77 meters and 𝑏 is equal to 36 meters. The length of the hypotenuse of this triangle gives the distance that Michael runs, which we’ll call 𝑑 subscript m. And so we have that 𝑑 subscript m is equal to the square root of 77 meters squared plus 36 meters squared. Taking the squares of these numbers, then under the square root, we get 5929 meters squared plus 1296 meters squared. These two numbers add to give 7225 meters squared. When we take the square root, we find that the distance Michael runs, 𝑑 subscript m, is equal to 85 meters.
Now that we know both Michael’s speed, which is 2.5 meters per second, and the distance that he runs, which is 85 meters, then we can use this formula here to calculate the time that he takes. We’ll call this time 𝑡 subscript m. And we have that 𝑡 subscript m is equal to the distance, 85 meters, divided by his speed, 2.5 meters per second. When we do the calculation, we find a time of 𝑡 subscript m equals 34 seconds.
Okay, We’re almost there now. But the question asks us for the difference in time at which the two children arrive. We know how long each child takes to get from the bottom left-hand corner to the top right-hand corner of the field. For Sofia, this time is 𝑡 subscript s equals 20 seconds, while for Michael, despite taking the more direct path, his time is longer at 𝑡 subscript m equals 34 seconds. To calculate the difference in arrival time, we subtract Sofia’s time from Michael’s time. We’ll call this time difference capital 𝑇 so that we have capital 𝑇 equals 𝑡 subscript m minus 𝑡 subscript s.
What we’re doing here is taking the time for which Michael is running and subtracting from that the time for which Sophia is also running. What we’re left with is the time for which Michael is still running after Sophia has already arrived at the top right-hand corner, i.e., the difference in their arrival times. Putting in the numbers, we have 𝑡 subscript m is equal to 34 seconds and 𝑡 subscript s is equal to 20 seconds. So the difference in their arrival times is 34 seconds minus 20 seconds. Doing this subtraction gives a result of 14 seconds.
So our answer to the question is that the difference in the time taken for Sophia and Michael to arrive at the top right-hand corner of the field is equal to 14 seconds.
