A car moved 150 meters east and then 225 meters north. Find the magnitude and direction of its displacement, rounding the angle to the nearest minute.
Let’s begin by recalling the four main compass points north, south, east, and west. If we let the unit vector 𝐢 hat equal one meter in the easterly direction and the unit vector 𝐣 hat correspond to one meter in the northerly direction, then we can write the displacement of the car in terms of these unit vectors. We are told that the car moves 150 meters east. This is equal to 150𝐢. It then travels 225 meters north, which is equal to 225𝐣. The displacement vector 𝐬 of the car is therefore equal to 150𝐢 plus 225𝐣.
We are asked to find the magnitude of this displacement. And we know that the magnitude of a vector is equal to the square root of the sum of the squares of its components. In this question, the magnitude of the displacement is equal to the square root of 150 squared plus 225 squared. Typing this into the calculator gives us 75 root 13. The magnitude of the displacement is 75 root 13 meters.
We can demonstrate this on our diagram with the use of a right-angled triangle where the base is equal to 150 and the height is 225. The direction of the displacement can be calculated by finding the angle 𝜃 between the horizontal and the direction of travel. Using our knowledge of right-angle trigonometry, we know that the tan of angle 𝜃 is equal to the opposite over the adjacent. In this question, the tan of angle 𝜃 is equal to 225 over 150.
Taking the inverse tangent of both sides of our equation, we have 𝜃 is equal to the inverse tan of 225 over 150. Typing this into the calculator, we get 56.3099 and so on degrees. As we are asked to give our answer to the nearest minute, this is equal to 56 degrees and 19 minutes. We can therefore conclude that the direction of the displacement of the car is 56 degrees and 19 minutes north of east.