Physics Question Help Please Quick 10 Points
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Physics Question Help Please Quick 10 Points

[From: ] [author: ] [Date: 11-09-08] [Hit: ]
400 mi away. He travels at a steady 52 mph. Beth leaves Los Angeles at 9:00 a.m. and drives a steady 60 mph. Who gets there first and how long does the first to arrive have to wait for the second?......
Alan leaves Los Angeles at 8:00 a.m. to drive to San Francisco, 400 mi away. He travels at a steady 52 mph. Beth leaves Los Angeles at 9:00 a.m. and drives a steady 60 mph. Who gets there first and how long does the first to arrive have to wait for the second?

Common sense would tell me that Beth gets there first, but what exact equation would I need to use. I have tried this problem several ways and keep getting around 1 minute for the answer, which is not correct. Any help would be appreciated. Thanks.

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distance = rate*time

Alan travels 400 miles at 52 miles per hour so 400 = 52*t 400/52 = 7 hours and 41.5 minutes

since he left at 8:00 am, he will arrive at 3:41:30pm

For Beth 400=60/t so 400/60 = 6 hours 40 minutes

since she left at 9:00am she will arrive first at 3:40 pm and will have to wait 1.5 minutes for Alan.

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Alan. Travel time = (400/52), = 7.692 hrs.
Beth = (400/60) = 6.667 hrs. So, Beth takes a total 7.667 hrs., measured from Alan's departure time.
Beth arrives first, by (7.692 - 7.667) = 0.025 hr., = (0.025 x 60) = 1.5 mins.
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