Enginpasa1 wrote:
In regards to the two object in the same direction, you reccomend that I SUBTRACT the rates. This baffles me! My approach is to apply the distance to the speed of body A and subtract that from the distance to the speed of body B. In other words, I would subtract the action (time-distance-rate) from eachother but only after treating them separately. Is there an example that you can provide to help explain this idea?
Sure - let's even look at one that doesn't involve traditional distance/rate/time.
Bob starts reading
War and Peace, a 1000 page book, at 9:00 a.m., at a rate of 50 pages per hour. Fred starts reading an identical copy of
War and Peace at 11:00 a.m., at a rate of 70 pages per hour. At what time will Bob and Fred be starting to read the same page of the book?
First, we need to figure out where Bob is when Fred starts reading. Bob has a 2 hour head start and is reading 50 pages per hour, so after 2 hours Bob will be 100 pages into the book. Therefore, Fred is 100 pages behind.
Fred reads at 70 pages per hour and Bob reads at 50 pages per hour. To find the relative page/hour rate, we SUBTRACT the individual rates. 70-50=20. Therefore, each hour Fred reads an additional 20 pages.
The distance that Fred needs to catch up is 100 pages and each hour Fred reads an extra 20 pages.
Time = distance/rate = 100/20 = 5 hours.
So, Bob and Fred will be on the same page 5 hours after Fred starts reading. 11:00 a.m. + 5 hours = 4:00 p.m.
