Q75P - University Physics with Modern Physics

Question

Two soccer players, Mia and Alice, are running as Alice passes the ball to Mia. Mia is running due north with a speed of 6.00m/s . The velocity of the ball relative to Mia is 5.00m/s in a direction $\mathbf{30}\mathbf{.}\mathbf{0}\mathbf{°}$ east of south. What are the magnitude and direction of the velocity of the ball relative to the ground?

Step-by-Step Solution

Answer

The magnitude of the velocity will be $v_{A/C}=3.01m/s$ and the direction will be $33.7°$  North of East.

Step 1: Introduction:

From the formula of the relative velocity of any object, if any object A is moving related to B and B is moving related to C, then the velocity of A related to C is,

 

 ${\mathit{v}}_{\mathbf{A}\mathbf{/}\mathbf{C}}\mathbf{=}{\mathit{v}}_{\mathbf{A}\mathbf{/}\mathbf{B}}\mathbf{+}{\mathit{v}}_{\mathbf{B}\mathbf{/}\mathbf{C}}$

Step 2: Given data:

For horizontal motion,

 

$$\begin{gathered} v_{A/B_h}=2.50 m/s \\ {v}_{B/C_h}=0.00 m/s \end{gathered}$$ 

 

For vertical motion,

                                                                                    

 $$\begin{gathered} v_{A/B_v}=-4.33 m/s \\ {v}_{B/C_v}=6.00 m/s \end{gathered}$$

Step 3: Magnitude and direction of velocity relative to the ground:

For horizontal motion,

 

$$\begin{gathered} v_{A/C_h}=v_{A/B_h}+v_{B/C_h} \\ =2.50 m/s+0.00 m/s \\ =2.50 m/s \end{gathered}$$ 

                                                                                    

For vertical motion,

                                                                                    

 $$\begin{gathered} v_{A/C_v}=v_{A/B_v}+v_{B/C_v} \\ =-4.33m/s+6.00 m/s \\ =1.67 m/s \end{gathered}$$

                                                                                                                                                             

Now the magnitude of velocity relative to the ground is define as below.

                                                                                    

$$\begin{gathered} v_{A/C}=\sqrt{{\left(v_{A/C_h}\right)}^2+{\left(v_{A/C_v}\right)}^2} \\ =\sqrt{{\left(2.50m/s\right)}^2+{\left(1.67m/s\right)}^2} \\ =\sqrt{6.25{\left(m/s\right)}^2+2.79{\left(m/s\right)}^2} \\ =3.01 m/s \end{gathered}$$ 

                                                                                   

And the direction of velocity relative to the ground is given by,

 

 $$\begin{gathered} \tan \theta =\frac{v_{A/C_v}}{v_{A/C_h}} \\ =\frac{1.67 m/s}{2.50 m/s} \\ \theta ={\tan }^{-1}\left(0.668\right) \\ =33.7° \end{gathered}$$

  

Direction will be  $33.7°$ North of East.