Problem 73 - University Physics with Modern Physics

Question

In a World Cup soccer match, Juan is running due north toward the goal with a speed of 8.00 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 37.0\(^\circ\) 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

Ball's velocity relative to Juan: 7.39 m/s at 77.6° east of north.

Step 1: Understand the Problem

We need to determine the velocity of the ball relative to Juan, who is running north. The ball's velocity relative to the ground has both northward and eastward components due to the angle of 37.0° east of north.

Step 2: Break Down the Ball's Velocity

Find the components of the ball's velocity:- Northward component: \[ v_{b,n} = 12.0 \times \cos(37.0^{\circ}) \approx 9.6 \text{ m/s} \] - Eastward component:\[ v_{b,e} = 12.0 \times \sin(37.0^{\circ}) \approx 7.2 \text{ m/s} \]

Step 3: Find Juan's Velocity Components

Juan is moving due north with 8.00 m/s, so his components are: - Northward: 8.00 m/s - Eastward: 0 m/s

Step 4: Calculate Relative Velocity Components

Subtract Juan's velocity components from the ball's to find the relative velocity:- Relative northward velocity:\[ v_{rel,n} = v_{b,n} - v_{j,n} = 9.6 - 8.0 = 1.6 \text{ m/s} \]- Relative eastward velocity:\[ v_{rel,e} = v_{b,e} - v_{j,e} = 7.2 - 0 = 7.2 \text{ m/s} \]

Step 5: Calculate Magnitude of Relative Velocity

Use the Pythagorean theorem:\[ v_{rel} = \sqrt{v_{rel,n}^2 + v_{rel,e}^2} = \sqrt{(1.6)^2 + (7.2)^2} \approx 7.39 \text{ m/s} \]

Step 6: Determine Direction of Relative Velocity

Find the direction angle \( \theta \) using the arctan of the eastward to northward component ratio:\[ \theta = \tan^{-1}\left(\frac{v_{rel,e}}{v_{rel,n}}\right) = \tan^{-1}\left(\frac{7.2}{1.6}\right) \approx 77.6^{\circ} \]This means the ball's velocity is \( 77.6^{\circ} \) east of north relative to Juan.

Key Concepts

Velocity Components

When dealing with motion, especially in two dimensions, it’s essential to break down velocity into its fundamental parts: velocity components. These components lie along the coordinate axes, which usually involve splitting into north-south and east-west directions for earth-related motions.
To better understand this, consider the ball moving at 12.0 m/s and 37.0° east of north. We cannot directly add this to Juan’s velocity because it’s not aligned with the same direction. Instead, we use trigonometry to split the ball's velocity into two components:

• Northward component: This prong of the velocity aligns with the direction Juan is running. Calculated as: \[ v_{b,n} = 12.0 \times \cos(37.0^{\circ}) \approx 9.6 \, \text{m/s} \]
• Eastward component: This represents how far the ball drifts to Juan’s right as it travels. Calculated as: \[ v_{b,e} = 12.0 \times \sin(37.0^{\circ}) \approx 7.2 \, \text{m/s} \]

Breaking motion into these parts helps visualize and solve problems involving movement in two or more directions seamlessly.

Pythagorean Theorem

The Pythagorean theorem is an essential mathematical principle that helps determine the magnitude of a resultant vector when dealing with right-angled triangles. When we talk about relative velocity, this theorem aids in calculating the combined effect of different motion components.
In our scenario, we aim to find the ball's speed relative to Juan. Having determined the relative northward and eastward components as 1.6 m/s (north) and 7.2 m/s (east), respectively, we employ the Pythagorean theorem. The theorem states:\[ c^{2} = a^{2} + b^{2} \] where *c* is the hypotenuse representing the final velocity. Plugging in the given values:
\[ v_{rel} = \sqrt{(1.6)^{2} + (7.2)^{2}} \approx 7.39 \, \text{m/s} \] This calculation tells us that, ignoring his own movement, Juan experiences the ball traveling at about 7.39 m/s in their relative frame of reference. The Pythagorean theorem is indispensable in physics whenever motion components along orthogonal directions need to be combined into a single resultant quantity.

Angle of Direction

Once you've dissected velocities into components and found the magnitude of relative velocity, the angle of direction helps complete the picture by determining how these components orient the resultant vector in a plane.
This is calculated using the tangent function and the arctan, or inverse tangent, which relates the eastward component with the northward component. In more straightforward terms, you find the angle *θ* by:

• Using the ratio of the eastward to northward velocities: \[ \theta = \tan^{-1}\left(\frac{v_{rel,e}}{v_{rel,n}}\right) = \tan^{-1}\left(\frac{7.2}{1.6}\right) \approx 77.6^{\circ} \]

This result tells us that Juan perceives the ball moving at an angle of 77.6° east of north. Understanding angle of direction is pivotal for providing detailed descriptions of motion trajectories, which can be critical in navigation, animations, or sports strategy analysis.