Q34E - University Physics with Modern Physics

Question

The nose of an ultralight plane is pointed due south, and its airspeed indicator shows 35m/s. The plane is in a 10m/s wind blowing toward the southwest relative to the earth. (a) In a vector-addition diagram, show the relationship of ${\stackrel{\mathbf{\rightharpoonup }}{\mathbf{v}}}_{\mathbf{P}\mathbf{/}\mathbf{E}}$ (the velocity of the plane relative to the earth) to the two given vectors. (b) Let x be east and y be north, and find the components of ${\stackrel{\mathbf{\rightharpoonup }}{\mathbf{v}}}_{\mathbf{P}\mathbf{/}\mathbf{E}}$. (c) Find the magnitude and direction of ${\stackrel{\mathbf{\rightharpoonup }}{\mathbf{v}}}_{\mathbf{P}\mathbf{/}\mathbf{E}}$.

Step-by-Step Solution

Answer

1. The relationship between v_P/E and given vectors are as below.
   

2. The components of v_P/E are $\left[-7.07\hat{i}-42\hat{j}\right]m/s$.

3. The magnitude and direction of velocity are 42.6m/s and 9.6^o.

Step 1: Identification of given data:

The given data can be listed below.

• The relative speed of plane is, v_P = 10m/s
• The wind speed indicator of plane is, v = 35m/s

Step 2: Concept/Significance of relative speed:

Relative velocity is defined as the velocity of an object relative to another observer. It is the time rate of change of relative position of one object with respect to another object.

 

Relative speed is the rate at which one thing moves or is at rest in comparison to another item that moves or is at rest.

Step 3: (a) Sketch a vector-addition diagram to describe the relationship of v ⇀ P / E (the velocity of the plane relative to the earth) to the two given vectors

The diagram of velocity using vector addition is given as below.

Step 4: (b) Determination of x be east and y be north, and find the components of v ⇀ P / E .

The velocity of windspeed relative to earth is given by,

$\begin{array}{rcl}{v}_{M/E}& =& -10\cos \left({45}^o\right)\hat{i}-10\sin \left({45}^o\right)\hat{j}\\ & =& \left(-7.07\hat{i}-7.07\hat{j}\right)m/s\end{array}$

The component of velocity of plane relative to earth is given by,

$v_{P/E}=v_{P/W}+v_{W/E}$

Substitute all the values in the above equation.

$\begin{array}{rcl}{v}_{P/E}& =& \left[-7.07\hat{i}-7.07\hat{j}-35\hat{j}\right]m/s\\ & =& \left[-7.07\hat{i}-42\hat{j}\right]m/s\end{array}$

Thus, the components of v_P/E are $\begin{array}{rcl}& & \left[-7.07\hat{i}-42\hat{j}\right]m/s\end{array}$.

Step 5: (c) Determination of the magnitude and direction of v ⇀ P / E .

The magnitude of velocity of plane relative to earth is given by,

$v_{PE}=\sqrt{v_x^2+v_y^2}$

Here, v_x is the x-component of velocity of plane and v_y is the y-component.

Substitute all the values in the above,

$\begin{array}{rcl}{v}_{PE}& =& \sqrt{{\left(-7.07m/s\right)}^2+{\left(-42m/s\right)}^2}\\ & =& \sqrt{49.9849{\left(m/s\right)}^2+1764{\left(m/s\right)}^2}\\ & =& 42.6m/s\end{array}$

The direction of the plane is given by,

${\theta }_2={\tan }^{-1}\left(\frac{v_y}{v_x}\right)$

Substitute all the values in the above,

$\begin{array}{rcl}{\theta }_2& =& {\tan }^{-1}\left(\frac{-7.07}{42}\right)\\ & =& 0.168\\ & =& 9.6^o\end{array}$

Hence, the magnitude and direction of velocity are 42.6m/s and 9.6^o.