# Question Video: Calculating Resultant Displacement Physics • 9th Grade

A vacationer stays in a tent on a campsite that is part of a landscape, as shown in the diagram. Two paths run from the campsite, one of which runs east to some shops. A sign at the campsite says that the shops are 100 m distant and that the other path is 130 m long and leads to some ruins. The vacationer walks to the shops and sees a sign there that says a garage is located 230 m to the north. The garage is 320 m west of the nearest town to the campsite. What is the magnitude of the displacement from the ruins to the town, to the nearest meter?

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### Video Transcript

A vacationer stays in a tent on a campsite that is part of a landscape, as shown in the diagram. Two paths run from the campsite, one of which runs east to some shops. A sign at the campsite says that the shops are 100 meters distant and that the other path is 130 meters long and leads to some ruins. The vacationer walks to the shops and sees a sign there that says a garage is located 230 meters to the north. The garage is 320 meters west of the nearest town to the campsite. What is the magnitude of the displacement from the ruins to the town to the nearest meter?

Okay, so we’re being asked to find the magnitude of the displacement from the ruins, which is this point here on our diagram, to the town, which is this point up here. This displacement vector can be represented by an arrow that starts with its tail at the position of the ruins and has its tip at the position of the town. The magnitude of the displacement is the length of the arrow. We can identify a right-angled triangle in our diagram which we’ve shown in blue. We know that this angle must be 90 degrees because we’re told that this path runs north. And we’re told that the town is east from the garage, so this road must run east.

The hypotenuse of this right-angled triangle is the displacement vector from the ruins to the town. And as for the other two sides, one of them is the length of the north path between the ruins and the garage, and the other is the main road which goes west from the town to the garage. We know that the road between the garage and the town is 320 meters long. But we don’t know the length of the north path between the ruins and the garage.

What we do know is that the garage is located north of the shops by a distance of 230 meters. This 230 meters must be equal to the distance from the ruins to the garage, which is this side of our triangle, plus the distance from the shops to the ruins. So if we label the distance from the shops to the ruins as 𝑥 one and the distance from the ruins to the garage as 𝑥 two, then we have that 230 meters is equal to 𝑥 one plus 𝑥 two.

We can also identify a second right-angled triangle in the diagram. Since we know that this path runs north and this one we can see runs east, then this angle must be 90 degrees. The reason that we’re busy looking for right-angled triangles in the diagram is so that we can use Pythagoras’s theorem to help us solve this problem. The magnitude of the displacement that we’re trying to find is the hypotenuse of a right-angled triangle in the diagram. And Pythagoras’s theorem connects the length of the hypotenuse of a right-angled triangle to the lengths of the other two sides.

Specifically, for a general right-angled triangle with a hypotenuse of length 𝑐 and other sides of lengths 𝑎 and 𝑏, Pythagoras’s theorem states that 𝑐 squared is equal to 𝑎 squared plus 𝑏 squared. If we take the square root of both sides of this equation, we get that the length of the hypotenuse 𝑐 is equal to the square root of 𝑎 squared plus 𝑏 squared. If we look at this triangle here in our diagram, we know that the hypotenuse is the magnitude of the displacement that we’re trying to find. And we can see that one of the other sides of the triangle has a length of 320 meters and the final side has an unknown length 𝑥 two.

So if we can work out the value of this unknown quantity 𝑥 two, then we can use it along with our value of 320 meters as our values of 𝑎 and 𝑏 in the right-hand side of our equation from Pythagoras’s theorem. To work out the value of 𝑥 two, we’re going to use this equation here. If we subtract 𝑥 one from both sides of the equation, then on the right-hand side the 𝑥 one and the minus 𝑥 one cancel each other out. And so we find that 𝑥 two is equal to 230 meters minus 𝑥 one. This means that if we can work out the value of 𝑥 one, then all we need to do is to subtract this value from 230 meters to get our value for 𝑥 two.

Now, in order to find 𝑥 one, we can notice that it’s one of the sides in our other right-angled triangle. And in this triangle we know that the length of the hypotenuse is 130 meters and the length of the other side is 100 meters. If we compare this triangle with our general triangle from Pythagoras’s theorem, then we can identify this 130-meter hypotenuse as our side 𝑐, this 100-meter horizontal length as the side 𝑎, and the length 𝑥 one as our side marked 𝑏. So in this equation for Pythagoras’s theorem, we know the values of both 𝑐 and 𝑎. And we want to find the value of the quantity 𝑏. This means that we want to rearrange the equation to make 𝑏 the subject.

To do this, we first subtract 𝑎 squared from both sides of the equation. On the right-hand side, the 𝑎 squared and the minus 𝑎 squared cancel out, which leaves us with 𝑏 squared is equal to 𝑐 squared minus 𝑎 squared. If we then take the square root of both sides of the equation, then on the left-hand side the square root of 𝑏 squared is simply 𝑏. And so we have that 𝑏 is equal to the square root of 𝑐 squared minus 𝑎 squared. Let’s now clear ourselves a bit of space so that we can substitute in 130 meters in place of 𝑐 and 100 meters in place of 𝑎 in this equation.

Okay, we’ve moved our equation connecting 𝑥 two and 𝑥 one over here on the left and we’re ready to substitute values into this equation. In place of 𝑏, we’ve got our side length of 𝑥 one and this is equal to the square root of the square of 130 meters minus the square of 100 meters. So under the square root, that’s the square of the hypotenuse, 𝑐, minus the square of the other side, 𝑎. The square of 130 meters minus the square of 100 meters works out as 6900 meters squared. Then evaluating the square root gives us that 𝑥 one is equal to 83.066 meters. These ellipses here are used to indicate that this result has further decimal places.

Now that we’ve got our value for 𝑥 one, we can use this equation over here to calculate 𝑥 two. Taking this value of 𝑥 one and substituting it into this equation, we have that 𝑥 two is equal to 230 meters minus 83.066 meters, which works out as 146.934 meters. If we now turn our attention to this triangle in our diagram, we see that we know the lengths of both of the non-hypotenuse sides. So if we substitute these two values in place of the 𝑎 and 𝑏 in this equation from Pythagoras’s theorem, then the value of 𝑐 that we’ll calculate will be the length of this hypotenuse. We’ve seen already that the length of this hypotenuse is equal to the magnitude of the displacement from the ruins to the town, which is exactly what we’re asked to find.

So let’s clear a little more space and substitute our values into this equation from Pythagoras’s theorem. The equation says that 𝑐 is equal to the square root of 𝑎 squared plus 𝑏 squared. And in this case, 𝑐 is the length of this hypotenuse here. And in place of the quantities 𝑎 and 𝑏, we’ve got 320 meters and 146.934 meters. The square of 320 meters plus the square of 146.934 meters works out as 123989.53 meters squared. Evaluating the square root gives a result for 𝑐 of 352.12 meters. Since we’re asked to give our answer to the nearest meter, we need to round our result for 𝑐. Rounding to the nearest meter gives us our answer that the magnitude of the displacement from the ruins to the town is 352 meters.

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