For this case we have to write the equation of the position of each one.
We have then:
For the robber:
[tex] r = 14t [/tex]
For the marshall :
[tex] r = 16 (t- (\frac{20}{60})) [/tex]
[tex] r = 16 (t-\frac{1}{3}) [/tex]
By the time the marshall reaches the robber we have:
[tex] 14t = 16 (t-\frac{1}{3}) [/tex]
From here, we clear the time.
We have then:
[tex] 16t - 14t = \frac{16}{3} [/tex]
[tex] 2t = \frac{16}{3} [/tex]
[tex] t =\frac{8}{3} [/tex]
[tex] t = 2.7 h [/tex]
Answer:
it takes the marshall to catch up to the robber about 2.7 hours.