Reference no: EM13121013
The equation for the distance of the room is: d=20+5t and the equation for the distance of Peter is d=vt =10t
Such that setting the two equations together we see that at t=4 seconds, the distance that Peter travelled is the same as the distance of the room 40ft.
However, here he only drops of one paper and he must go back to pick up the rest, can he catch up again? The two equations are never equal again.
Here is the Problem:
Peter has finished some written papers and must now cross the room to turn them in to Mariel.
But Peter can only carry one page at a time across the 20 ft distance from him to Mariel and the distance he must cross increases by 5 ft at the end of every second.
Peter can move 10 ft per second, and the algebra papers take a negligible amount of time to pick up (imagine that it is instantaneous) or put down.
(ie. at 0secs the distance is 20ft, at 1sec, the distance is 25ft, at three seconds the distance is 30ft. But his velocity is always 10ft/second)
QUESTION:
1. How many pages of his write-up can peter get to Mariel in the 5 minutes of class remaining? 5 minutes = 5x60 =300 seconds
2. How wide is the room at each point Peter i) either reaches Mariel or ii)returns to his own place?
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