A student observes an airplane fly overhead with a constant velocity
parallel to the x-axis at a height of 7.60 × 103 m. At time t = 0 the airplane is directly
above the student, the vector P~
o = 7.60 × 103ˆj m describes the position on the airplane
with respect to the student. At time t = 30.0 s, the position vector has changed to
P~
30 =
�
8.04 × 103ˆi + 7.60 × 103ˆj
�
m as shown in the figure below. Determine the magnitude
and orientation of the airplane’s position vector at t = 45.0 s.
To determine the magnitude and orientation of the airplane's position vector at t = 45.0 s, we can use the given position vectors and apply vector subtraction.
Let's denote the position vector at time t = 0 as P~o and the position vector at time t = 30 as P~30. Using these vectors, we can calculate the displacement vector from t = 0 to t = 30.
Let's subtract P~o from P~30 to get the displacement vector:
ΔP~ = P~30 - P~o
= (8.04 × 103ˆi + 7.60 × 103ˆj) - (0ˆi + 7.60 × 103ˆj)
= (8.04 × 103ˆi + 0ˆj)
Now, we need to find the position vector at t = 45.0 s, which is given by P~45 = P~o + ΔP~.
P~45 = P~o + ΔP~
= (0ˆi + 7.60 × 103ˆj) + (8.04 × 103ˆi + 0ˆj)
= 8.04 × 103ˆi + 7.60 × 103ˆj
The magnitude of the position vector P~45 is given by the Pythagorean theorem:
|P~45| = √((8.04 × 103)^2 + (7.60 × 103)^2)
Now, calculating the magnitude:
|P~45| = √(64.3216 × 106 + 57.76 × 106)
= √(122.0816 × 106)
= √(122.0816) × 103
≈ 11.05 × 103 m
Therefore, the magnitude of the airplane's position vector at t = 45.0 s is approximately 11.05 × 103 m.
To determine the orientation of the position vector, we can use trigonometry. The orientation can be represented by the angle θ, where θ is the angle between the position vector and the positive x-axis.
θ = tan^(-1)(y/x) = tan^(-1)((7.60 × 103)/(8.04 × 103))
Now, calculating the angle:
θ = tan^(-1)(0.9443)
≈ 43.19°
Therefore, the orientation of the airplane's position vector at t = 45.0 s is approximately 43.19° with respect to the positive x-axis.