Question

A soccer ball rests on the field at the location \(x=5.0 \mathrm{~m}\). Two players run along the same straight line toward the ball. Their equations of motion are as follows: $$ \begin{aligned} &x_{1}=-8.2 \mathrm{~m}+(4.2 \mathrm{~m} / \mathrm{s}) t \\ &x_{2}=-7.3 \mathrm{~m}+(3.9 \mathrm{~m} / \mathrm{s}) t \end{aligned} $$ (a) Which player is closer to the ball at \(t=0\) ? (b) At what time does one player pass the other player? (c) What is the location of the players when one passes the other?

Short answer

(a) Player 2, (b) 3 seconds, (c) 4.4 meters.

Step-by-step solution

Evaluate Initial Positions (a)

To find which player is closer to the ball at \(t = 0\), substitute \(t = 0\) into both equations. For player 1: \(x_1 = -8.2 \space \text{m} + (4.2 \space \text{m/s}) \cdot 0 = -8.2 \space \text{m}\). Similarly, for player 2: \(x_2 = -7.3 \space \text{m} + (3.9 \space \text{m/s}) \cdot 0 = -7.3 \space \text{m}\). Now, we compare their distances to the ball's position (5.0 meters). Player 1 is \(5.0 - (-8.2) = 13.2 \space \text{m}\) away, and player 2 is \(5.0 - (-7.3) = 12.3 \space \text{m}\) away. Player 2 is closer.

Set Equations Equal (b)

To find when one player passes the other, set their equations equal: \(-8.2 + 4.2t = -7.3 + 3.9t\). Solve for \(t\) to find the time they meet.

Solve for t (b)

\(-8.2 + 4.2t = -7.3 + 3.9t\) simplifies to \(-8.2 + 4.2t - 3.9t = -7.3\). This becomes \(-8.2 + 0.3t = -7.3\). Solving for \(t\), we get \(0.3t = -7.3 + 8.2 = 0.9\). Thus, \(t = \frac{0.9}{0.3} = 3\) seconds.

Determine Location at Meeting Time (c)

Substitute \(t = 3\) seconds into either player's equation to find their location at the time they pass each other. Using player 1's equation: \(x_1 = -8.2 + 4.2 \times 3 = -8.2 + 12.6 = 4.4 \space \text{m}\). Therefore, both players are at 4.4 meters.

Key concepts

Equations of Motion

In physics, equations of motion describe the mathematical relationships between an object’s position, velocity, and acceleration over time. For the two players running towards the soccer ball, their equations of motion show their starting points and rates of movement.
Each player has an equation in the form:

• Player 1: \( x_1 = -8.2 \, \text{m} + (4.2 \, \text{m/s}) t \)
• Player 2: \( x_2 = -7.3 \, \text{m} + (3.9 \, \text{m/s}) t \)

Here, the initial positions are \

Relative Position

Relative position describes the location of one object concerning another. In this context, we're interested in how each player's position compares to the soccer ball's position.
At \(t = 0\), we find the players have initial positions of \(-8.2 \, \text{m}\) and \(-7.3 \, \text{m}\), respectively. These positions are negative because they are behind the ball's position on the field, which is at +5.0 m.

• To find their distance from the ball, consider the soccer ball at +5.0 m and calculate:
  • Player 1: \(5.0 - (-8.2) = 13.2 \, \text{m}\)
  • Player 2: \(5.0 - (-7.3) = 12.3 \, \text{m}\)

This tells us that Player 2 starts closer to the soccer ball.

Meeting Point

A meeting point is where two objects moving towards each other reach the same position. For our players, they meet when their positions, based on time, become equal.
To find this moment, set the two equations of motion equal to each other:

• \(-8.2 + 4.2t = -7.3 + 3.9t\)

Rearrange and solve for \(t\):

• \(-8.2 + 0.3t = -7.3\)
• \(0.3t = 0.9\)
• \(t = \frac{0.9}{0.3} = 3\, \text{seconds}\)

Both players meet at 3 seconds, and substituting \(t = 3\) into either equation provides the location:

• Substitute into Player 1's equation: \(-8.2 + 4.2 \times 3 = 4.4 \, \text{m}\)

Therefore, at \(t = 3\) seconds, both players are at 4.4 meters.