You drive 7.50 km in a straight line in a direction 15° east of north. Find the distances – Free B8
Displacement Vector Components: 7.50 km at 15° East of North
Question:
You drive 7.50 km in a straight line in a direction 15.0° east of north.
(a) Find the distances you would have to drive straight east and then straight north to arrive at the same point. (This is equivalent to finding the components of the displacement along the east and north directions.)
- a) 1.94 km and 7.24 km
- b) 3.75 km and 6.50 km
- c) 6.50 km and 3.75 km
- d) 7.24 km and 1.94 km
Answer:
Correct option: a) 1.94 km and 7.24 km
Detailed Explanation:
We are given a displacement of 7.50 km at an angle of 15.0° east of north. To resolve this vector into its east (x-direction) and north (y-direction) components, we apply basic trigonometry.
Step 1: Identify the Trigonometric Relationships
x (east) = 7.50 × sin(15°)
y (north) = 7.50 × cos(15°)
y (north) = 7.50 × cos(15°)
Using the known values:
- sin(15°) ≈ 0.2588
- cos(15°) ≈ 0.9659
Step 2: Calculate Each Component
x ≈ 7.50 × 0.2588 = 1.94 km
y ≈ 7.50 × 0.9659 = 7.24 km
y ≈ 7.50 × 0.9659 = 7.24 km
This means to reach the same final point, you could alternatively drive:
- 1.94 km east, and then
- 7.24 km north.
Conclusion:
The correct pair of displacement components is:
East = 1.94 km
North = 7.24 km
North = 7.24 km
Thus, the correct answer is option (a).