Question

An airplane A is flying horizontally due east at a speed of $400km/hr$. Passengers in A observe another airplane B moving perpendicular to direction of motion at A. Airplane B is actually moving in a direction $30°$north of east in the same horizontal plane as shown in the figure. Determine the velocity of B.

Answer 1

Step 1: Given

Velocity of plane A: $v_A=400km/hr$

Velocity vector for plane A: ${\overrightarrow{V}}_A=400\hat{i}$

Velocity of plane B is $v_B$.

Velocity vector for plane B is ${\overrightarrow{V}}_B$.

Angle made by plane B in north-east direction: $\theta =30°$

Step 2: Formula Used

${\overrightarrow{V}}_x=v_x\cos \theta \hat{i}+v_y\sin \theta \overrightarrow{j}$

${\overrightarrow{V}}_{x/y}={\overrightarrow{V}}_x-{\overrightarrow{V}}_y$

Step 3: Find the velocity of plane B

Find the velocity vector of plane B by substituting values in the formula

$$\begin{gathered} {\overrightarrow{V}}_B=v_B\cos \theta \hat{i}+v_B\sin \theta \overrightarrow{j} \\ =v_B\cos 30\hat{i}+v_B\sin 30\overrightarrow{j} \\ =v_B\frac{\sqrt{3}}{2}\hat{i}+v_B\frac{1}{2}\overrightarrow{j} \end{gathered}$$

Subtract the velocity vector of A from that of B

$$\begin{gathered} {\overrightarrow{V}}_{B/A}={\overrightarrow{V}}_B-{\overrightarrow{V}}_A \\ =\left(v_B\frac{\sqrt{3}}{2}\hat{i}+v_B\frac{1}{2}\hat{j}\right)-400\hat{i} \\ =\left(v_B\frac{\sqrt{3}}{2}-400\right)\hat{i}+v_B\frac{1}{2}\hat{j} \end{gathered}$$

Calculate the velocity of plane B by equating the component along north to 0.

$$\begin{gathered} v_B\frac{\sqrt{3}}{2}-400=0 \\ \Rightarrow v_B\frac{\sqrt{3}}{2}=400 \\ \Rightarrow v_B=400\times \frac{2}{\sqrt{3}} \\ \Rightarrow v_B=\frac{800}{\sqrt{3}}km/hr \end{gathered}$$

Substitute the value in ${\overrightarrow{V}}_B=v_B\frac{\sqrt{3}}{2}\hat{i}+v_B\frac{1}{2}\overrightarrow{j}$.

$$\begin{gathered} {\overrightarrow{V}}_B=v_B\frac{\sqrt{3}}{2}\hat{i}+v_B\frac{1}{2}\overrightarrow{j} \\ =\frac{800}{\sqrt{3}}\frac{\sqrt{3}}{2}\hat{i}+\frac{800}{\sqrt{3}}\frac{1}{2}\overrightarrow{j} \\ =400\hat{i}+\frac{400}{\sqrt{3}}2\overrightarrow{j} \end{gathered}$$

Hence, the velocity of plane B is $\frac{800}{\sqrt{3}}km/hr$ and the velocity vector is $400\hat{i}+\frac{400}{\sqrt{3}}2\overrightarrow{j}$.