Homework Help: math/physics

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

During a race, one runner runs at a 6 min/mi pace and another runner runs at a 7 min/mi pace. The fast runner passes the slow runner. After a 2 second delay, the slow runner initiates an acceleration, catching up with the fast runner in 968 feet. Give this acceleration in ft/sec^2.

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.

No one has answered this question yet.

Answer this Question

Related Questions

Physics - Runner A is running at a constant rate of 2.3 m/s when he passes ...
physics - In an 8.00 race, one runner runs at a steady 11.8 and another runs at ...
Physics - At an instant, Runner 'A' runs at 6m/s with an acceleration of...
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 ...
math - A long-distance runner runs 3 miles south and then 7 miles east. How far ...
math - A marathon runner ran the first 5 mi. in 32.5 min. If she continues ...
math - 11. Runner A crosses the starting line of a marathon and runs at an ...
Math - Runner A crosses the starting line of a marathon and runs at an average ...
math - one runner runs 4.75 miles in one hour and runner 2 runs 4 miles in one ...

For Further Reading

First Name:
School Subject:
Answer: