About average speed please
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About average speed please

[From: ] [author: ] [Date: 11-04-24] [Hit: ]
The time for Melvin and Jason to get this far apart is 1,080/5=216minutes.Total distance between towns is Stephens distance+Melvins distance at the time they meet=21,600meters+16,200meters=37,800met… or 37.......
Stephen left Town A and walked towards Town B at a speed of 100m/min. At the same time, Jason and Melvin started from Town B and walked towards Town A at a speed of 80m/min and 75m/min respectively. If Stephen met Melvin six minutes after passing Jason, find the distance between Town A and Town B.

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This question should be in the Math section. Let t= the time of walking, then vt=distance of walking. Stephen's distance from Jason to Melvin=100x6=600meters
Melvin's distance from Stephen's meeting Jason to Stephen's meeting Melvin=75x6=450meters
Jason's distance from Stephen's meeting Jason to Stephen's meeting Melvin=80x6=480meters
Total distance between Melvin and Jason is 600+450+30=1,080meters at the time Stephen meets Melvin
The time for Melvin and Jason to get this far apart is 1,080/5=216minutes.
Total distance between towns is Stephen's distance+Melvin's distance at the time they meet=21,600meters+16,200meters=37,800met… or 37.8km
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