Question

1. you are standing on the ground at the origin of a coordinate system. an airplane flies over you with constant velocity parallel to the x - axis and at a fixed height of 7.60×10³ m. at time t = 0, the airplane is directly above you so that the vector leading from you to it is $vec{p}_{0}=7.60×10^{3}hat{j}$. at t = 30.0 s, the position vector leading from you to the airplane is $vec{p}_{30}=(8.04×10^{3}hat{i}+7.60×10^{3}hat{j})$ as suggested in the figure. determine the magnitude and orientation of the airplanes position vector at t = 45.0 s.

Explanation:

Step1: Find the x - component velocity

The x - component of the position vector changes from 0 at \(t = 0\) to \(x_{30}=8.04\times 10^{3}\text{ m}\) at \(t = 30.0\text{ s}\). Using the formula \(v_x=\frac{\Delta x}{\Delta t}\), we have \(v_x=\frac{8.04\times 10^{3}-0}{30.0}=268\text{ m/s}\).

Step2: Find the x - component of the position vector at \(t = 45.0\text{ s}\)

Using the formula \(x = v_x t\), with \(v_x = 268\text{ m/s}\) and \(t = 45.0\text{ s}\), we get \(x=268\times45.0 = 12060\text{ m}\). The y - component remains constant \(y = 7.60\times 10^{3}\text{ m}\). So the position vector \(\vec{P}_{45}=12060\hat{i}+7.60\times 10^{3}\hat{j}\).

Step3: Calculate the magnitude of the position vector

The magnitude of a vector \(\vec{A}=A_x\hat{i}+A_y\hat{j}\) is given by \(|\vec{A}|=\sqrt{A_x^{2}+A_y^{2}}\). For \(\vec{P}_{45}\), \(|\vec{P}_{45}|=\sqrt{(12060)^{2}+(7.60\times 10^{3})^{2}}=\sqrt{145443600 + 57760000}=\sqrt{203203600}\approx14255\text{ m}\).

Step4: Calculate the orientation of the position vector

The orientation \(\theta\) of the vector \(\vec{P}_{45}\) with respect to the x - axis is given by \(\tan\theta=\frac{y}{x}\). So \(\tan\theta=\frac{7.60\times 10^{3}}{12060}\approx0.63\), and \(\theta=\arctan(0.63)\approx 32.2^{\circ}\) above the x - axis.

Answer:

Magnitude: \(14255\text{ m}\), Orientation: \(32.2^{\circ}\) above the x - axis