While following the directions on a treasure map, a pirate walks 38.8 m north, then turns and walks 5.8 m east.

What is the magnitude of the single straight-line displacement that the pirate could have taken to reach the treasure?
Answer in units of m
At what angle with the north would he have to walk?
Answer in units of ◦

a. D^2 = X^2 + Y^2

D^2 = 5.8^2 + 38.8^2 = 1539.1
D = 39.2 m.

b. tan A = Y/X = 38.8/5.8 = 6.68966
A = 81.5o, CCW.
A = 90-81.5 = 8.5o East of North.

To find the magnitude of the single straight-line displacement, we can use the Pythagorean theorem. The displacement is the hypotenuse of a right triangle formed by the north and east directions.

1. Square the magnitude of the north displacement: (38.8)^2 = 1504.64 m^2
2. Square the magnitude of the east displacement: (5.8)^2 = 33.64 m^2
3. Add the two squared magnitudes: 1504.64 + 33.64 = 1538.28 m^2
4. Take the square root of the total: √1538.28 ≈ 39.2 m

The magnitude of the single straight-line displacement is approximately 39.2 meters.

To find the angle with the north, we can use the inverse tangent function.

1. Calculate the angle tangent: tan^(-1)(5.8/38.8) ≈ 8.92°

Therefore, the pirate would need to walk at an angle of approximately 8.92° with respect to the north direction to reach the treasure.