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A starts from place a by bicycle and B starts from place B by foot. After two hours of meeting each other, it is known that the speed of a is three times that of B, which is more than 2km / h A. B. the distance between the two places is 44km (1) Find the speed of a and B; (2) After meeting, they continue to move forward in their respective directions. How long does it take for a to go back to B and catch up with B?
If B's speed is a km / h, then a's speed is 3A + 2 km / h
(a+3a+2)*2=44
4a+2=22
4a=20
A = 5 km / h
Speed of B = 5 km / h
A's speed = 5 * 3 + 2 = 17 km / h
It takes 44 / 17 hours for a to arrive at B
At this time, B walked 5 × 44 / 17 = 220 / 17 km
44-220 / 17 = 528 / 17 km from a
Speed difference = 17-5 = 12 km / h
It takes (220 / 17) / 12 = 55 / 51 hours for a to catch up with B
At this time, B has a total of 5 × (2 + 55 / 51) = 785 / 51 = 15.40
Both Party A and Party B have driven from place a to place B at the speed of 60km per hour. At 8:20, the distance from place a is twice that from place B,
At 8:26, when the ratio of the distance between a and B is 3:2, the time of a is 3:2
Suppose: at 8:20, the time for a is t, and the time for B is d,
60T/60D=2
[60（T+6/60)]/[60(D+6/60)=3/2
T = 1 / 5 hours, that is 12 minutes
D = 1 / 10 hours, i.e. 6 minutes
So a starts at 8:08
The departure time of Party B is 8:14
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