After robbing a bank in Dodge City, a robber gallops off at 13 mi/h. 20 minutes later, the marshall leaves to pursue the robber at 16 mi/h. How long (in hours) does it take the marshall to catch up to the robber?

let the time taken by the robber when he is caught be t hrs

distance he covered = 13t miles

time taken by the marshall to catch him = t - 1/3
distance traveled by him = 16(t - 1/3)

but 13t = 16(t - 1/3)
13t = 16t - 16/3
-3t = -16/3
t = 16/9 hrs

time of marshall = t - 1/3
= 16/9 - 1/3 = 13/9 hrs or 1 hour and appr 27 minutes

To find out how long it takes for the marshall to catch up to the robber, we need to determine the distance traveled by both the robber and the marshall.

Let's start by converting the time of the delay of the marshall from minutes to hours. Since there are 60 minutes in an hour, 20 minutes is equal to 20/60 = 1/3 hour.

The robber's head start can be calculated as the distance traveled by him during this delay period. As the robber is traveling at 13 mi/h, multiplying his speed by the delay time gives us: 13 mi/h * (1/3) h = 13/3 mi.

Now, we can determine the time it takes for the marshall to catch up to the robber. Let's assume this time is 't' hours.

Since the marshall is traveling at 16 mi/h and the robber's speed is constant, their relative speed (for the marshall catching up) can be calculated by subtracting the robber's speed from the marshall's speed: 16 mi/h - 13 mi/h = 3 mi/h.

Now, we set up an equation to find the time 't' it takes for the marshall to catch up:

Relative Speed * Time = Distance
3 mi/h * t = 13/3 mi

Solving for 't', we get:
t = (13/3 mi) / (3 mi/h)

Simplifying the equation, we have:
t = (13/3) / 3
t = 13/9

Therefore, it takes the marshall approximately 13/9 hours to catch up to the robber.