Monday

November 24, 2014

November 24, 2014

Posted by **victoria** on Sunday, January 2, 2011 at 1:06pm.

I have figured out that after the 2 seconds when the fast runner passes the slow runner, there are 4.19047 feet between the runners. I got this by:

5280 ft/360 sec = x ft/2 sec and

5280 ft/420 sec = x ft/2 sec then finding the difference in the distance the runners had travelled over those 2 seconds. However, I can't figure out what to do now.

**Answer this Question**

**Related Questions**

Physics - Runner A is running at a constant rate of 2.3 m/s when he passes ...

Math - Runner A crosses the starting line of a marathon and runs at an average ...

physics - Runner A is running at a constant rate of 2.3 m/s when he passes ...

Physics - At an instant, Runner 'A' runs at 6m/s with an acceleration of 2m/s^2...

physics - In an 8.00 race, one runner runs at a steady 11.8 and another runs at ...

math - Two runners are competing in a race. One runner averages a pace of 7 ...

math - Two runners are competing in a race. One runner averages a pace of 7 ...

Physics Help!!!! - A runner in a relay race runs 20m north, turns around and ...

Math - Runner A crosses the starting line of a marathon and runs at an average ...

math - 11. Runner A crosses the starting line of a marathon and runs at an ...