# Question Video: Resultant Motion and Force Physics • 9th Grade

A bird flies along a line that displaces it 450 m east and 350 m north of its starting point, as in the diagram. What angle must the bird turn toward the west to change direction and fly directly north? Give your answer to the nearest degree.

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### Video Transcript

A bird flies along a line that displaces it 450 meters east and 350 meters north of its starting point, as shown in the diagram. What angle must the bird turn toward the west to change direction and fly directly north? Give your answer to the nearest degree.

All right, so we have a diagram where the cardinal directions are labeled such that east is toward the right and north is toward the top. The blue line represents the displacement vector of the bird. And the two black lines represent the vertical and horizontal components of this displacement. As stated in the question, these are 350 meters north and 450 meters east, respectively. The question asks us what angle the bird must turn to change its direction and fly directly towards the north.

To help us figure out this angle, let’s draw the bird’s northward trajectory on the diagram. This magenta line represents a northward trajectory since it points directly up. On the other hand, if the bird were to continue along its current trajectory, it would follow this blue dotted line. So in order to change direction and fly directly north, the bird has to turn through this angle between the two trajectories.

This is the angle we need to calculate. But since we don’t know anything else about the dotted or magenta vectors, let’s try to find another part of our diagram with the same angle. The magenta vector and the northward part of the displacement both point directly north, so they’re parallel. But this means that the bird’s trajectory is a straight line intersecting two parallel lines. So corresponding angles at the two intersections have the same measure.

In our diagram, this angle between the northern component of the displacement and the displacement itself corresponds to the angle we drew before, since both of these angles are to the right of one of the parallel lines and above the displacement line. So if we can find the measure of this angle, we’ll know the angle that the bird needs to turn.

Now, recall that by adding the horizontal and vertical components of a vector, we get the vector itself. Therefore, if we draw 450 meters to the east between the head of the northern vector and the head of the displacement vector, we’ll get a right triangle formed by the two components and the displacement. These two vectors form a right angle because north and east are perpendicular. Let’s call our angle 𝜃.

Now, recall that the tangent of an angle is equal to the length of the opposite side divided by the length of the adjacent side. So tan of 𝜃 is equal to 450 meters divided by 350 meters. Meters divided by meters is just one. So the right-hand side of this equation is just a number without units. Now, we can take the inverse tangent of both sides to find this angle. Plugging into a calculator 𝜃 is equal to arctan of 450 divided by 350, we get 𝜃 is very close to 52.125 degrees.

Now, remember, this is the same angle that we were looking for to answer the question. So all we need to do is round 𝜃 to the nearest degree. Looking to the first digit right of the decimal point, one is less than five, so two rounds to two. So, to the nearest degree, the bird needs to turn 52 degrees toward the west to fly directly north.