how long does it take for one to catch another?

Question

I have used many equations to try to solve this problem but I still am far from the correct answer.
Here is the problem:

A theif runs out of a store he has robbed, and runs down the street at 10km/hr. 6 minutes later the guard wakes up and starts chasing the robber at 14km/hr.
A. How far does the theif get in 6 minutes? - i got 1km.
B. How long does it take for the guard to catch up with the theif?
C. How far from the store are they when he is caught?
D. What is the guard's average velocity from when the theif leaves until he is caught?

Answer 1

Part a is correct.

For the guard

x = vg*(t), vg =14 km/hr

for the thief

x = vt+ 1km


Set the two equal vt+ 1 = vg*t ---> t = 1/(vg - v) = 1/(4) =.25 hrs = 15 minutes

x = vg*(.25 hrs) = 14*.25 = 3.5 km

Guard sleeps for 6 min. so t = 15 + 6 =21 min. = 0.35 hrs

ave speed = 3.5 km/.35 hrs = 10 km/hr

Answer 2

EASY

THE DUDE RUNS 10 KM- SO 1 KM
SECURITY DUDE RUNS 14KM
SO HE IS CATCHING DUDE AT 4KM AN HOUR
HE NEEDS TO GET 1 KM
AND SINCE DUDE IS 1KM AWAY, IT WILL TAKE 15 MINUTES FOR SECURITY TO CATCH DUDE
14/4= 3.5 KM AWAY FROM STORE
DONT KNOW ABOUT D- SORRY

CAN I PLEASE PRETTY PLEASE WITH A CHERRY ON TOP HAVE THE POINTS
I