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作者: 1a2b3c4z 時間: 16-3-7 22:00 標題: 問數學

A jogging trail was 3km long. Mary and Steve's average speed was 3.2km/h & 2.8km/h respectively. They started at the starting point, but Mary started 3min after Steve. How long until he caught up with Steve ?

作者: Ryan_baba 時間: 16-3-7 23:55

將他們的速度分別轉成m/s就好易計

作者: jane-wang 時間: 16-3-8 08:12 標題: 回覆：1a2b3c4z 的帖子

几年级的数学题？学方程没有？

作者: 81545 時間: 16-3-8 10:06

回覆 1a2b3c4z 的帖子

1) Steve's walking distance after 3 min = 2800/60 x 3 min = 140 metres
2) Let the minutes be X,

3200X/60 = 2800X/60 +140
400X/60 = 140
X = 21

作者: yeesumyin 時間: 16-3-8 13:34

Because Steve had run for 2.8km x 3/60 = 0.14km, so the distance between Steve and Mary was 0.14km after 3 mins. As Mary could run 0.4km more than Steve in an hour, so the formula is:
2.8km x 3/60 / 0.4 x 60
= 21min.

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