A person walks in the following pattern: 4.8 km north, then 1.3 km west, and finally 1.5 km south. Construct the vector diagram that represents this motion and from it judge how far and in what direction a bird would fly in a straight line from the same starting point to the same final point. (Choose east as the +x direction.)

the distance d is found by

d^2 = (4.8-1.5)^2 + 1.3^2

How do you find the direction (in degrees). I said 21.5 degrees originally, but that was wrong

Your angle of 21.5° is right, but if you draw the diagram, you will see that the bird is roughly NW, so it must fly roughly SE.

So, it is really -21.5° (relative to the x-axis), or a heading of 111.5° from due North.

To construct the vector diagram, we need to represent the different displacements of the person as vectors. Let's consider the north direction as the positive y-axis and the east direction as the positive x-axis.

1. Start by drawing a horizontal line to represent the x-axis (east).
2. From the origin (starting point), draw a vertical line upward to represent the north displacement of 4.8 km. Label this vector as A.
3. From the end point of vector A, draw a horizontal line leftward to represent the west displacement of 1.3 km. Label this vector as B.
4. From the end point of vector B, draw a vertical line downward to represent the south displacement of 1.5 km. Label this vector as C.
5. Connect the starting point (origin) to the end point of vector C to create the bird's straight-line path. Label this vector as D.

To find the magnitude and direction of vector D (the bird's path), we can use the Pythagorean theorem and trigonometry.

Magnitude:
The length of vector D can be found using the Pythagorean theorem:
|D| = √(x² + y²)

Direction:
The direction of vector D can be found using the inverse tangent (tan⁻¹) function:
θ = tan⁻¹(y/x)

Let's calculate the magnitude and direction of vector D.

Magnitude:
|D| = √(1.3² + (4.8 - 1.5)²)
|D| = √(1.69 + 9.41)
|D| = √11.1
|D| ≈ 3.33 km

Direction:
θ = tan⁻¹((4.8 - 1.5)/1.3)
θ = tan⁻¹(3.3/1.3)
θ ≈ 68.7°

Therefore, the bird would fly approximately 3.33 km in a straight line at an angle of about 68.7° east of north from the same starting point to the same final point.