The boy initially travels 3 km north and then 4 km east. To determine the direction and displacement, we can use vector addition.
The 3 km north is represented by a vector pointing straight up (towards the north) with a magnitude of 3 km. Let's call this vector A.
The 4 km east is represented by a vector pointing to the right (east) with a magnitude of 4 km. Let's call this vector B.
To find the direction and displacement, we need to find the resultant vector by adding vectors A and B.
Adding vectors A and B using vector addition, we can draw a right triangle with sides of length 3 km and 4 km. The hypotenuse of this triangle will represent the resultant vector, which is the displacement.
Using the Pythagorean theorem, we can find the magnitude of the resultant vector:
Resultant magnitude = sqrt((3 km)^2 + (4 km)^2) = sqrt(9 km^2 + 16 km^2) = sqrt(25 km^2) = 5 km.
So, the magnitude of the resultant vector (displacement) is 5 km.
To find the direction, we can use trigonometry. The angle between the resultant vector and the north direction can be found using the inverse tangent function:
Angle = arctan((opposite side) / (adjacent side)) = arctan(3 km / 4 km) ≈ 36.87 degrees.
Therefore, the direction of the displacement is approximately 36.87 degrees east of north.
In summary, the boy's displacement is 5 km at an angle of approximately 36.87 degrees east of north.