# Video: Calculating Displacement Vector

The F-35B is a short takeoff and vertical landing fighter jet. An F-35B takes off vertically upward to a height of 20.00 m, while still facing horizontally. The fighter then follows a flight path that is angled at 30.00° above a line parallel to the ground, 20.00 m vertically above ground. The fighter flies for a distance of 20.00 km along this trajectory. What is the fighter’s final displacement?

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### Video Transcript

The F-35B is a short takeoff and vertical landing fighter jet. An F-35B takes off vertically upward to a height of 20.00 meters, while still facing horizontally. The fighter then follows a flight path that is angled at 30.00 degrees above a line parallel to the ground, 20.00 meters vertically above ground. The fighter flies for a distance of 20.00 kilometers along this trajectory. What is the fighter’s final displacement?

We can call the displacement of the fighter after these maneuvers 𝑑, where we understand that 𝑑 is a vector with both magnitude and direction. To begin on our solution, let’s draw out a sketch of this F-35B’s motion. We’re told that the jet fighter rises vertically a distance we’ve called ℎ of 20.00 meters. And then begins to move at an angle of 30.00 degrees to the horizontal along a path of length 20.00 kilometers. The fighter’s displacement 𝑑 is the vector that connects its initial location on the ground with its final location at the end of its 20.00 kilometer of leg. It’s that vector that we want to solve for based on this given information. We can start by orienting the motion of our fighter to a set of coordinate axes. We’ll position a pair of 𝑥- and 𝑦-coordinate axes, so that their origin, the place where they cross, is the place where our jet fighter began its journey. This means we can write our displacement vector 𝑑 in the following form. A component, we’ll call it 𝑑 sub 𝑥, in the 𝑖-direction plus a component, we’ll call it 𝑑 sub 𝑦, in the 𝑗-direction.

So now, our task is to set out and solve for 𝑑 sub 𝑥 and 𝑑 sub 𝑦. Looking first at 𝑑 sub 𝑥, we see that that value, the 𝑥-component of the jet fighter’s motion, will be equal to 𝑙 times the cos of 30.00 degrees. We know that 𝑙 is given as 20.00 kilometers or 20.00 times 10 to the third meters. When we multiply that by the cos of 30.00 degrees, we find a value of 1.732 times 10 to the fourth meters. That’s 𝑑 sub 𝑥. Now we want to solve for 𝑑 sub 𝑦, the vertical displacement of the jet fighter. Looking at our diagram, we see that 𝑑 sub 𝑦 is equal to ℎ plus 𝑙 times the sin of 30.00 degrees. Plugging in for ℎ and 𝑙, we once again use a value of 20.00 times 10 to the third meters for 𝑙. And ℎ, we’re given as 20.00 meters. With these values entered on our calculator, we find that 𝑑 sub 𝑦, to four significant figures, is 1.002 times 10 to the fourth meters. We insert this value for 𝑑 sub 𝑦 in our expression for displacement, 𝑑. And this gives us the final form for the displacement of the jet fighter, 1.732 times 10 to the fourth meters in the 𝑖-direction and 1.002 times 10 to the fourth meters in the 𝑗-direction.