Question

An aeroplane A is flying horizontally due east at a speed of 400km/hr. Passengers in A, observe another aeroplane B moving perpendicular to direction of motion at A. aeroplane B is actually moving in a direction of 30 degree north of east in the same horizontal plane . Determine the velocity of B.

Answer 1

Answer:

the correct answer is B

Answer 2

Given:

An aeroplane A is flying horizontally due east at a speed of 400km/hr. Passengers in A, observe another aeroplane B moving perpendicular to direction of motion at A. aeroplane B is actually moving in a direction of 30 degree north of east in the same horizontal plane .

To Find:

Determine the velocity of B?

Solution:

here, it is given that-

An aeroplane A is flying horizontally due east at a speed of 400km/hr.

so, $V_{A}= 400 km/hr$ , $\[{\vec V_A} = 400\hat i\]$

Also, passengers in A, observe another aeroplane B moving perpendicular to direction of motion of A making an angle of 30 degree north of east in same horizontal plane.

therefore, velocity of B is given by-

$\[{\vec V_B} = {V_B}\cos (\pi /6)\hat i + {V_B}\sin (\pi /6)\hat j\]$

this implies;

$\[\begin{gathered} {{\vec V}_{B/A}} = {{\vec V}_B} - {{\vec V}_A} \hfill \\ {{\vec V}_B} = \left( {{v_B}\dfrac{{\sqrt 3 }}{2}\hat i + \dfrac{{{v_b}}}{2}\hat j} \right) - 400\hat i \hfill \\ {{\vec V}_{B/A}} = \left( {{v_B}\dfrac{{\sqrt 3 }}{2}\hat i - 400} \right)\hat i + \dfrac{{{v_B}}}{2}\hat j \hfill \\ \end{gathered} \]$

$\[\begin{gathered} \sqrt 3 {V_B} - 800 = 0 \hfill \\ {V_B} = \frac{{800}}{{\sqrt 3 }} \hfill \\ \end{gathered} \]$

Hence, velocity vector of B is given by;

$\[{V_B} = 400\hat i + \frac{{400}}{{\sqrt 3 }}\hat j\]$