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The distance between a and B is 21km. After a starts for 1.5h, B starts again. A is in the back, B is in the front, walking in the same direction. A rides 8km every hour, B walks every hour 5km, how many hours after a starts to catch up with B?
Set off for X hours and catch up with B
8x-5（x-1.5）=21
3x+7.5=21
3x=13.5
x=4.5
A: after 4.5 hours, a will catch up with B
Xiaoming and Xiaoliang are 4km apart. Xiaoming starts 0.2h first, Xiaoliang starts again, Xiaoming is behind and Xiaoliang is in front. They are walking in the same direction. Xiaoming's speed is 8km / h
Xiaoliang's speed is 6km / h. Xiaoming catches up with Xiaoliang a few hours after starting
Set X hours after departure to synchronize with Xiao Ming
6X+4=8（0.2+X）
6X+4=1.6+8x
2x=2.4
x=1.2
1.2+0.2=1.4
A: Xiao Ming catches up with Xiao Liang after 1.4 hours of departure
I hope it can be adopted
A car drives from place a to place B at the speed of 60km / h, and it returns from place B to place a at the speed of 40km / h. The average speed of the car is calculated
Let the distance between a and B be s, the time from a to B be T1, and the time from B to a be T2
t1=s/60；t2=s/40
Average speed v = 2S / (T1 + T2) = 2S / (s / 60 + S / 40) = 2 / (1 / 60 + 1 / 40) = 2 / [(2 + 3) / 120] = (2 * 120) / 5 = 48km / h
So the average speed of the car is 48km / h
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