Question

Student bikes to school by traveling first dN = 1.10 miles north, then dW = 0.300 miles west, and finally dS = 0.200 miles south.

Part A
Take the north direction as the positive y direction and east as positive x. The origin is still where the student starts biking. Let d⃗ N be the displacement vector corresponding to the first leg of the student's trip. Express d⃗ N in component form.

Answer 1

The displacement vector d⟧N corresponding to the first leg of the student's trip north is in component form as (0, 1.10) miles, where the x-component is 0 (no east/west movement) and the y-component is 1.10 (northward movement).

Step-by-step explanation:

The student bikes to school by traveling first dN = 1.10 miles north, then dW = 0.300 miles west, and finally dS = 0.200 miles south. To express the displacement vector d⟧N corresponding to the first leg of the student's trip in component form, you need to use the coordinate system provided with north being the positive y direction and east as the positive x direction. Since the student is moving north and there is no east/west component in this leg of the trip, the x component is 0.

Therefore, the displacement vector d⟧N in component form is (0, 1.10) miles. Here, the x-component represents the east/west direction, and the y-component represents the north/south direction. The first value is the x-component which is 0 because there is no displacement in the east/west direction for this part of the trip. The second value is the y-component which is 1.10, representing 1.10 miles north.

Answer 2

Taking the vertical component of the displacement
1.1 - 0.2 = 0.9 mile
The horizontal component of the displacement
-0.3 mile

The magnitude of the displacement is
√[ (0.9)² + (-0.3) ] = 0.95 mile

The direction is
θ = tan-1 (-0.3/0.9)
θ = 161.57 degrees.