Question

Question: Two spectators at a soccer game see, and a moment later hear, the ball being kicked on the playing field. The time delay for spectator Ais, 0.23 s and the spectator Bit is 0.12s. Sight lines from the spectators to the player kicking the ball meet at an angle of. How far are (a) spectatorand (b) spectatorfrom the player? (c) How far are the spectators from each other?

Short answer

Answer

1. Spectator A is 79 m far from the player.
2. Spectator B is 41 m far from the player.

3. The spectators are 89 far from each other.

Step-by-step solution

Given data 

1. The time delay for spectator A is$\Delta t_A=0.23\text{s}$.
2. The time delay for spectator B is$\Delta t_B=0.12\text{s}$.
3. The slight line from the two spectators to the player kicking the ball meet at an angle of 90⁰ .

Determining the concept

The time for sound to travel from the kicker to a spectator is given by,

$t_{sound}=\frac{d}{v}$

The time required for light to travel the same distance dis given by,

$t_{light}=\frac{d}{c}$

Here, c is the speed of light, d is the distance and v is speed of sound.

(a) Determine how far is spectator A

The delay between seeing and hearing the kick is,

$$\begin{gathered} \Delta t=t_{sound}-t_{light} \\ \Delta t=\frac{d}{v}-\frac{d}{c} \end{gathered}$$

But, v<<c, so delay can be approximated by,

$$\begin{gathered} \Delta t=\frac{d}{v} \\ d=v\Delta t \end{gathered}$$

Thus, the distance from the kicker to spectator A is,

$d_A=v\Delta t_A$

Substitute 343 m/s for v and 0.23 s for $∆t_a$into the above equation

$$\begin{gathered} d_A=343\times 0.23 \\ =79\text{m} \end{gathered}$$

Hence, spectator A is 79m far from the player.

(b) Determine how far is spectator B

Similarly, the distance from the kicker to spectator is,

$d_B=v\Delta t_B$

Substitute 343 m/s for v and 0. 12s for $∆t_b$into the above equation

$$\begin{gathered} d_B=343\times 0.12 \\ =41 \text{m} \end{gathered}$$

Hence, spectator B is 41m far from the player.

(c) Determining the distance of spectators from each other

The line from the kicker to each spectator and from one spectator to the other, form a right angle with the line joining the spectators as the hypotenuse. So, the distance between the spectators is,

$D=\sqrt{{d_A}^2+{d_B}^2}$

Substitute 79 m for d_A and 41 for d_b into the above equation,

$$\begin{gathered} D=\sqrt{{79}^2+{41}^2} \\ =89\text{m} \end{gathered}$$

Hence, the spectators are far from each other.