Video: Addition of Three Vectors in Two Dimensions in a Real-World Context

A cyclist rides 5.0 km east, then 10.0 km at an angle 20° west of north. From this point, she rides 8.0 km west. What is the final displacement from where the cyclist started? Consider east to correspond to positive horizontal displacement and north to correspond to positive vertical displacement.

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Video Transcript

A cyclist rides 5.0 kilometres east, then 10.0 kilometres at an angle 20 degrees west of north. From this point, she rides 8.0 kilometres west. What is the final displacement from where the cyclist started? Consider east to correspond to positive horizontal displacement and north to correspond to positive vertical displacement.

Let’s call the displacement we want to solve for from the cyclist’s start to end point 𝑑. And we can start off by drawing a bird’s-eye view or aerial view of the cyclist’s motion. The diagram of the cyclist’s journey reveals that there are three legs. Starting at the origin, the cyclist goes 5.0 kilometres east, then 10.0 kilometres in a direction 20 degrees west of north, then 8.0 kilometres due west.

The cyclist’s overall displacement is the difference in position from her start point at the origin to her endpoint; it’s that displacement we want to solve for. Since displacement is a vector, meaning it has magnitude and direction, we want to know the cyclist’s displacement in terms of both her E direction or 𝐢 component motion as well as her north-south displacement or in the 𝐣 direction.

Let’s call the three segments of the cyclist’s journey 𝑠 one, 𝑠 two, and 𝑠 three. For each one, we want to figure out the 𝐢 component and the 𝐣 component of that segment. Let’s start with 𝑠 one, the leg of the journey going 5.0 kilometres east. Since east is in the positive 𝐢 direction, the 𝐢 component of this leg is plus 5.0 kilometres and the 𝐣 component is zero.

Moving on to 𝑠 two, the 𝐢 component of 𝑠 two is solved using trigonometry. If we draw a horizontal line from the end of the 𝑠 two line segment to the dotted vertical line, then that dotted vertical line and the line we’ve just drawn form a right angle. So there is a right triangle, whose hypotenuse is 10.0 kilometres in magnitude. The 𝐢 component of that triangle is equal to that magnitude multiplied by the sine of 20 degrees. And because that leg of the triangle is pointing in the western direction, it will have a negative value according to our coordinate plane. That value is negative 10.0 kilometres times the sine of 20 degrees.

Similarly, the 𝐣 component of 𝑠 two is equal to that magnitude 10.0 kilometres times the cosine of 20 degrees. In this case, the value is positive because it’s in the northern direction, which we’ve defined as positive. And finally, we move on to 𝑠 three. The 𝐢 component of 𝑠 three is negative because 𝑠 three is motioned to the west and it’s negative 8.0 kilometres. There is no 𝐣 component to 𝑠 three; that component is zero.

Now our task is to add up the 𝐢 and 𝐣 columns separately. When we add the values in our 𝐢 column, we get a result of negative 6.4 kilometres and in the 𝐣 column, we get a value of positive 9.4 kilometres. Therefore, we can write the displacement of the cyclist 𝑑 as negative 6.4 𝐢 plus 9.4 𝐣 kilometres. This is the total displacement of the cyclist from start to finish.

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