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Question from Jay:

There is a police officer in pursuit of suspect. They suspect is driving on a straight road traveling at 79 mph, the officer is 233 yds away, traveling at 92 mph, the suspects car 5 ft long and the officers car is 5.3 ft long, how long will it take to have the front of the officers car, equal to the front of the suspects car down to the tenth of a second.

Hi Jay,

In 1940 a science fiction writer, Robert Heinlein wrote a book titled "The Roads Must Roll".The story is set in the near future, when "roadtowns" (wide rapidly moving passenger platforms similar to moving sidewalks, but reaching speeds of 100 mph) have replaced highways and railways as the dominant transportation method in the United States. I think of Heinlein's roads when I see this type of problem.

Suppose you have a road moving at 79 miles per hour and the suspect's car is stationary on the road.The police car is also on the road travelling at 13 mph. To someone not on the road it looks like the suspect is travelling at 79 mph and the police car is travelling at 79 + 13 = 92 mph, but to the police officer, the suspect's car is stationary and his car is travelling at 13 mph. Travelling at 13 mph how long will it take the police car to overtake the suspect's car?

You need to take care here since you have a mixture of units. I would express the 13 mph in feet per second and the 233 yards in feet. When the problem says that "the officer is 233 yards away" I assume that to mean the front bumper of the police car is 233 yards behind the rear bumper of the suspect's car. The police car then has to travel 233 yards (expressed in feet) + 5 feet to bring his front bumper in line with the other car's front bumper. I think the length of the police car is not important.

Write back if you need any more assistance,
Harley

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