math/physics
posted by victoria .
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.
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