Concentric Semi-Circular Running Tracks Analysis

The problem describes five concentric, adjacent, semi-circular running tracks. We are given the mean length and width of each track, and the angle subtended by the innermost track. We need to find the angle subtended by the outermost track.

For an arc of a circle, the length ($L$) is related to the radius ($r$) and the angle subtended ($\theta$) by the formula:

\[L = r \theta\]

where $\theta$ is in radians.

We are given that the mean length of each track is 100 m and the width of each track is 2.5 m.

Let's label the tracks from 1 (innermost) to 5 (outermost).

For the innermost track (Track 1):

• Mean length, $L_1 = 100$ m
• Angle subtended, $\theta_1 = 1.00$ rad

Using the formula $L_1 = r_1 \theta_1$, where $r_1$ is the mean radius of the innermost track:

\[100 \text{ m} = r_1 \times 1.00 \text{ rad}\]

So, the mean radius of the innermost track is $r_1 = 100$ m.

The tracks are adjacent and each has a width $w = 2.5$ m. The mean radius of a track is the radius halfway across its width. If the inner edge radius of Track 1 is $R_{inner,1}$ and the outer edge radius is $R_{outer,1}$, then $r_1 = R_{inner,1} + w/2$. Also, $R_{outer,1} = R_{inner,1} + w$.

Since the tracks are adjacent, the outer edge of track $i$ is the inner edge of track $i+1$. The mean radius of track $i+1$ is $r_{i+1} = R_{inner,i+1} + w/2$. Since $R_{inner,i+1} = R_{outer,i} = R_{inner,i} + w$, we have $r_{i+1} = (R_{inner,i} + w) + w/2 = R_{inner,i} + 3w/2$.

Alternatively, and more simply, the mean radius of track $i$ is $r_i$. The outer edge of track $i$ is at radius $r_i + w/2$. This is the inner edge of track $i+1$. The mean radius of track $i+1$ is then $(r_i + w/2) + w/2 = r_i + w$.

So, the mean radii of the tracks are:

• $r_1 = 100$ m
• $r_2 = r_1 + w = 100 + 2.5 = 102.5$ m
• $r_3 = r_2 + w = 102.5 + 2.5 = 105.0$ m
• $r_4 = r_3 + w = 105.0 + 2.5 = 107.5$ m
• $r_5 = r_4 + w = 107.5 + 2.5 = 110.0$ m

The outermost track is Track 5. Its mean radius is $r_5 = 110.0$ m. We know its mean length is also $L_5 = 100$ m.

Let $\theta_5$ be the angle subtended by the outermost track (Track 5). Using the formula $L_5 = r_5 \theta_5$:

\[100 \text{ m} = 110.0 \text{ m} \times \theta_5\]

Now, we solve for $\theta_5$:

\[\theta_5 = \frac{100}{110.0} \text{ rad} = \frac{10}{11} \text{ rad}\]

Calculating the decimal value:

\[\theta_5 = \frac{10}{11} \approx 0.90909... \text{ rad}\]

Comparing this value with the given options, 0.90909... rad is closest to 0.91 rad.

Let's summarize the mean radii and check the relationship:

As the radius increases for the outer tracks, the angle subtended must decrease for the mean length to remain constant at 100 m.

The calculated angle for the outermost track is approximately 0.9091 rad, which rounds to 0.91 rad.

Final Answer for Outermost Track Angle

The angle subtended by the outermost track is approximately 0.91 rad.