Solution: Let R car speed be Vr and T car speed be vt = 0.13m/s.

(1) The second encounter between two cars shall meet the following requirements:

(Vr+Vt)*24=3*2

The imported data can be obtained in the following ways:

Vr=0. 12 (m/s)

(2) Because the cars are moving at a constant speed, the time required for the first 10 meeting of the two cars is:

24*5= 120 (seconds)

During this period, the distance traveled by R car is:

120 * 0.12 =14.4 (m)

Every 3 meters, the speed of the R car turns once and returns to the A place. After 120s, r turns 14.4/3=4 weeks+2.4m..

Therefore, the R train still needs to go through (3-2.4)/0. 12=5(s) to get back to a place.