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Solving the problem of two primary schools' mathematical Olympic circular travel
1、? Analysis: When A and B met for the first time, they only walked half a circle, that is, half a circle. Half a lap, B is gone 100 meters. When they met for the second time, they * * * walked one and a half laps. B can walk 100 meters for every half lap, and three and a half laps for a half lap, which means B walked three laps 100 meters, which is 300 meters. According to "A walked 60 meters a week before the second meeting", we can know that A walked a circle and a half with a difference of 60 meters, and A walked a circle and a half with a difference of 60 meters, indicating that B walked more than 60 meters and walked a half circle. B MINUS 60 meters from 300 meters is the length of half a circle. The length of a half circle is known, and the length of a week is easy to find.

Solution: (100× 3-60 )× 2 = 480 (m)

A:? The circumference of the circular field is 480 meters.

2、? Analysis: Three encounters show that they walked three 400 meters, the distance was 1200 meters, the meeting time was 8 minutes, and the sum of their speeds was 150 meters. According to "A walks 0. 1 meter more than B every second", we can know that A walks 6 meters more than B, so we can calculate the speeds of A and B.. When the velocities of A and B are calculated, other problems can be easily solved.

Solution: 400×3÷8= 150 (m)

0. 1 m/s = 6m/min

(150-6)÷2=72 (m)

72+6 = 78 (meter)

400× 2 ÷150× 78-400 = 416-400 =16 (m)

A:? The shortest distance between their second meeting point and point A along the runway is16m.