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A and B are 15km apart, and they are walking at the speed of 2.5km/h and 5km / h respectively, while the dog with a is 7.5km/h To B, the dog runs to a immediately after meeting B, runs to B when meeting a, and runs to a immediately after meeting B Until a and B meet, find the dog's distance. Use the equation solution
Let's assume that the dog has traveled XKM
x/7.5=15/（2.5+7.5）
The solution is x = 15
The distance between a and B is 12km. Now they are facing each other at the speed of 2.5km/h and 3.5km/h respectively. At the same time, a's dog is running at the speed of 10km / h
B, run to a immediately after meeting B, and run to B again after reaching A. until a and B meet, find the distance of the dog
The distance a dog walks is the product of the speed of the dog and the time it takes for a and B to meet
It takes 12 / (2.5 + 3.5) = 2 hours for Party A and Party B to meet
The distance a dog walks is 20 km
Party A and Party B are on the same road. Party A travels 5km per hour and Party B 7km per hour. Party A passes through place a at 12 noon and Party B passes through place a at 2 pm,
When can b catch up with a?
Linear equation of one variable
14-12 = 2 (hours)
2 × 3 ÷ (5-3) = 3 (hours)
3 + 2 = 5 (point)
B overtakes a at five
Let's catch up with a in X hours
3x+2×3=5x
6=2x
3=x
x=3
3 + 2 = 5 (point)
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