# A plane traveling with an air speed of 300km/hr [35 degrees W of N] relative to the wind. A wind...

## Question:

A plane traveling with an air speed of {eq}300 \ km/hr \ [35 ^o W \ \text{of} \ N] {/eq} relative to the wind. A wind is blowing at {eq}75 \ km/hr \ [E] {/eq} relative to the ground. Determine the velocity of the plane relative to the ground.

## Resultant vector

Vectors are quantities that contain both magnitude and direction. To solve this problem, we apply the vector addition or the sum of two vectors which is called the resultant vector. The equation that we will use is given below

\begin{align} V &= \sqrt{\sum v_x^2 + \sum v_y^2}\\ \end{align}

Given:

• Velocity of the plane {eq}v_1 = 300\ km/hr{/eq}
• Angle of the plane {eq}\theta_1 = -35 ^{\circ}{/eq} W of N
• Velocity of the wind {eq}v_2 = 75\ km/hr{/eq}
• Angle of the plane {eq}\theta_2 = 0{/eq} E

The velocity of the plane relative to the ground is just the resultant of the two velocities which is

\begin{align} V &= \sqrt{\sum v_x^2 + \sum v_y^2}\\ V &= \sqrt{\sum (v_1cos\theta_1 + v_2cos\theta_2)^2 + (v_1sin\theta_1 + v_2sin\theta_2)^2}\\ V &= \sqrt{((300\ km/hr)cos(-35) + (75\ km/hr)cos(0))^2 + ((300\ km/hr)sin(-35) + (75\ km/hr)sin(0))^2}\\ V &= \boxed{363.99\ km/hr} \end{align}

The direction is

\begin{align} tan\theta &= \frac{\sum v_y^2}{\sum v_x^2}\\ \theta &= tan^{-1} \left( \frac{((300\ km/hr)sin(-35) + (75\ km/hr)sin(0))}{((300\ km/hr)cos(-35) + (75\ km/hr)cos(0))} \right)\\ \theta &= \boxed{-61.79} \end{align}

The velocity of the plane is {eq}363.99\ km/hr{/eq} at {eq}61.79^{\circ}{/eq} W of N 