The road to El Chapo

Classical Mechanics Level 5

Recently, El Chapo, the leader of Mexico's Sinaloa cartel, was in the news for his incredible escape from the high security Altiplano prison. Over the course of months, members of his cartel dug a tunnel 30 ft underground from a decoy warehouse to the space directly below the shower of his prison cell, almost one mile in length!

To dig such a tunnel, one must remove dirt as one goes along, and bring it back outside of the tunnel. To accomplish this engineering feat, the workers made a rail system with a modified motorcycle which could hold approximately $0.5\ \text{m}^3$ of dirt per trip. If the rail bike moves at 5 m/s, and it takes the workers 30 min to remove dirt and fill it into the railbike, how long (in days) would it take them to dig the tunnel?

Details

The tunnel is 1500 m long, by 1.7 m tall, by 0.75 m wide.
The workers are active 24 hrs per day, every day, until the tunnel is finished.
The workers extend the track as they go, and this takes a negligible amount of time.

The answer is 92.9653.

1 solution

Chew-Seong Cheong
Jul 31, 2015

The length, height and width of the tunnel are $L = 1500$ m, $h = 1.7$ m and $w = 0.75$ m. Since a rail-bike load is $V = 0.5$ m $^3$ , each trip is a dig of length,

$l = \dfrac{V}{hw} = \dfrac{0.5}{1.7\times 0.75}$

The number of digs or trips necessary, $n = \dfrac{L}{l} = \dfrac{1500\times 1.7 \times 0.75}{0.5} = 3825$

The time taken to remove and fill in dirt $n$ trips, $t_r = 30n = 30 \times 3825 = 114750$ min

The total distance traveled in $n$ trips $D = \displaystyle \sum_{k=1}^n {2kl} = n(n+1)l = 5739000$ m

Since the rail bike moves at $v = 5$ m/s, the time taken to traveled $D$ , $t_t = \dfrac{D}{v} = \dfrac{5739000}{5} = 1147800$ s $=19130$ min

Total time spent: $t = t_r + t_t = \dfrac{114750 + 19130}{60\times 24} = \boxed{92.972}$ days

The road to El Chapo

The answer is 92.9653.

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