A and B start from a to B at the same time, and return immediately after arriving at B. when they return, their respective speeds increase by 20%. 1.5 hours after departure, the two vehicles meet. It is known that when B arrives at B, a still has 1 / 5 of the distance from a to ab. how many hours does it take a car to go back and forth between ab?

A and B start from a to B at the same time, and return immediately after arriving at B. when they return, their respective speeds increase by 20%. 1.5 hours after departure, the two vehicles meet. It is known that when B arrives at B, a still has 1 / 5 of the distance from a to ab. how many hours does it take a car to go back and forth between ab?

Suppose that the speed of car a is a, the speed of car B is B, and the distance between car a and car B is s
The analysis shows that a is faster than B. A has gone S + 4S / 5, and B has finished s, so there are some problems
S/a + (4S/5)/(1.2a) = S/b
So B = 3A / 5
Party A and Party B meet in 1.5 hours, yes
S-1.5b=(1.5-S/a)*1.2a
So there is
S / a = 27 / 22 (this is the time to go to Cheng a)
The return time is 5 / 6 of the return time
So it takes (1 + 5 / 6) * 27 / 22 = 27 / 12 hours for car a to go back and forth between ab,
That is 2 hours and 15 minutes
There are two situations
If the speed of car a is x and that of car B is y, then the speed of car a is 1.2x
The speed of car B is 1.2y, and the distance between a and B is 5S
Situation 1:
If a is faster than B, then B is faster
Formula 1: (5S / y) = (5S / x) + (4S / 1.2x)
Equation 2: 1.5y + 1.2x {1.5 - (5S / x)} = 5S
The solution is (s / x) = 0.245
The round trip time of car a between AB is (5S / x) + (5S / 1.2x) = 9.17 (s / x) = 2.25 hours
Case 2:
If B is faster than a, there will be
Formula 1: (5S / y) = (s / x)
Equation 2: 1.5x + 1.2y {1.5 - (5S / y)} = 5S
The solution is (s / x) = 0.927
The round trip time of car a between AB is (5S / x) + (5S / 1.2x) = 9.17 (s / x) = 8.5 hours
Note: the equal sign in the process of solving the problem is approximate equal sign
The answer upstairs is really speechless. I don't know who is fast in analyzing a and B, and how to firmly judge a upstairs
Faster than B. The Watchtower owner should consider the problem comprehensively
It is known that when B arrives at B, a still has 1 / 5 distance from a (return)
A is faster than B
A and B meet before B arrives at B. A returns to meet B after ab
Let a take x from a to B at the original speed
A is still 1 / 5 of the distance from a to B, which can be decomposed into the original speed from a to B, the time is x, plus the return walk
(1-1 / 5) of AB distance is 4 / 5, after the speed is increased by 20%, then the time is 4 / 5x divided by 120%, which is equal to 2 / 3x
The time for Party A and Party B to go is x + 2 / 3x = 5 / 3x hours
In other words, when B arrives at B, the time it takes is 5 / 3x, and the speed is naturally 3 / 5 of a's
After arriving at B, they return immediately. When they return, their speed increases by 20%. 1.5 hours after departure, the two cars meet
If the speed of a is 1, the speed of B is 3 / 5
According to the original speed, the time required for a to go from a to B is x, and the distance is x ab
1.5-x distance (1.5-x) * 1.2
B 1.5 hours walking distance 1.5 * 3 / 5 = 0.9 distance and ab
Then x = (1.5-x) 1.2 + 0.9 x = 27 / 22 hours
If the return speed of a is increased by 20%, it will only take 5 / 6 of the original time
How many hours does it take for car a to and fro between ab? 27 / 22 * (1 + 5 / 6) = 2 and 1 / 4 = 2.25 hours
Two hours, 15 minutes