A boy running in a northern direction later changes direction and started moving towards the Eastern direaction. If the distance he covered in the northern direaction 5km and in the Eastern direaction is 4km,determine a; the magnitude of his resultant.b; it's direaction
"towards" the Eastern direction? Why not just say East?
√(5^2 + 4^2) = √41 km @ NθE where
tanθ = 4/5
d = sqrt(25+16) = sqrt(41) = 6.4 approximately
tan angle north of east = 5/4
so angle = 51.3 degrees N of E
Which is 90 - 51.3 = 38.7deg E of N or compass bearing of boy from start
To solve this problem, we can use the Pythagorean theorem and trigonometry.
Step 1: Visualize the boy's movement on a 2D coordinate system. Let's assume his starting point is at the origin (0, 0).
Step 2: Calculate the magnitude of the resultant using the Pythagorean theorem:
Resultant magnitude (a) = √((distance north)² + (distance east)²)
= √((5km)² + (4km)²)
= √(25km² + 16km²)
= √(41km²)
≈ 6.40 km
Therefore, the magnitude of the resultant is approximately 6.40 km (a ≈ 6.40 km).
Step 3: Determine the direction of the resultant using trigonometry:
Let θ be the angle between the resultant and the north direction.
tan(θ) = (distance east) / (distance north)
tan(θ) = (4km) / (5km)
θ = tan^(-1)(4/5)
θ ≈ 38.66°
Since the boy started running in the northern direction and then moved towards the eastern direction, the angle between the resultant and the north direction is 38.66 degrees.
Therefore, the magnitude of the resultant is 6.40 km (a ≈ 6.40 km), and its direction is approximately 38.66 degrees east of north (b ≈ 38.66°).