Q73P - University Physics with Modern Physics

Question

In a World Cup soccer match, Juan is running due north toward the goal with a speed of 8.0 m/s  relative to the ground. A teammate passes the ball to him. The ball has a speed of 12.0 m/s and is moving in a direction $\mathbf{37}\mathbf{.}\mathbf{0}\mathbf{°}$ east of north, relative to the ground. What are the magnitude and direction of the ball’s velocity relative to Juan?

Step-by-Step Solution

Answer

The ball is moving at 7.39m/s in the direction $12.37°$ , relative to Juan.

Step 1: Relative velocity

From the formula of relative velocity of any object,

 

If any object P is moving related to A and A is moving related to E, then the velocity of P related to E is,

${\mathit{V}}_{\mathbf{P}\mathbf{/}\mathbf{E}}\mathbf{=}{\mathit{V}}_{\mathbf{P}\mathbf{/}\mathbf{A}}\mathbf{+}{\mathit{V}}_{\mathbf{A}\mathbf{/}\mathbf{E}}$
  

 Here, ${\mathit{V}}_{\mathbf{P}\mathbf{/}\mathbf{E}}\mathbf{,}{\mathit{V}}_{\mathbf{P}\mathbf{/}\mathbf{A}}\mathbf{,}{\mathit{V}}_{\mathbf{A}\mathbf{/}\mathbf{E}}$ are the, velocity of P related to E , velocity of P related to A , and velocity of A related to E respectively.

Step 2: Given

• The velocity of Juan relative to ground: ${\overrightarrow{V}}_{J/G}=8.0m/s$ due north.
• The velocity of the ball relative to ground:  ${\overrightarrow{V}}_{B/G}=12.0m/s 37.0°$ east of north. 

Step 3: Magnitude and direction of velocity of ball related to Jaun

The velocity of the ball with respect to Juan  ${\overrightarrow{V}}_{B/G}$ can be calculated using the relation-

 ${\overrightarrow{V}}_{B/G}={\overrightarrow{V}}_{B/J}+{\overrightarrow{V}}_{J/G}$

 

Splitting the magnitude of vectors in above relation into components along X and Y-axis. Components along X-axis are shown using the subscript X and the components along Y are shown using the subscript Y. thus the components of magnitude of velocity of ball, relative to juan, are-

 

$$\begin{gathered} V_{B/J_x}=V_{B/G_x} \\ =V_{B/G}\sin 37.0° \\ =\left(12.0 m/s\right)\sin 37.0° \\ =7.22m/s \\ {V}_{B/J_y}=V_{B/G_y}+V_{J/G_y} \\ =V_{B/G}\cos 37.0°+\left(8 m/s\right) \\ =\left(12 m/s\right)\cos 37.0°+\left(8 m/s\right) \\ =17.58 m/s \end{gathered}$$ 

  

The magnitude of velocity of ball relative to Juan can be obtained as-

 

$$\begin{gathered} V_{B/J}=\sqrt{{\left(V_{B/J_x}\right)}^2+{\left(V_{B/J_y}\right)}^2} \\ {V}_{B/J}=\sqrt{{\left(7.222m/s\right)}^2+{\left(17.58m/s\right)}^2} \\ {V}_{B/J}=19m/s \end{gathered}$$ 

                                                                                                                                                                                

The direction of velocity of ball relative to Juan can be obtained as-

 $$\begin{gathered} \tan \theta \text{'}=\frac{V_{B/J}}{V_{B/J_x}} \\ \tan \theta \text{'}=\frac{17.58m/s}{7.22m/s} \\ \theta \text{'}=67.67° \end{gathered}$$                                                                                          

  

Here,  $\theta \text{'}$ is the angle made by the  ${\overrightarrow{V}}_{B/J}$ from X-axis.

 

So, the magnitude and direction is 19 m/s and $67.67°$ north of east.