Answer Choices- 6, 7, 8, 14

Question

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Peter ran at 4 m/s. Paul started at the same spot 6 sec. later and chased Peter at 7 m/s. Paul needed __ seconds to catch Peter.

APatel Jan 17, 2019

ElectricPavlov · Answer 1 · 2019-01-17T04:38:22

Peter gets a 4 m/s x 6 s = 24 meter head start.....when do their positions equal?

24m + 4 m/s * x s = 7 m/s * x s

24 + 4x = 7x

24 = 3x

x = 8seconds they will be at the same point ( 24 m + 4 m/s * 8 sec = 56 meters)

CPhill · Answer 2 · 2019-01-17T04:39:37

Every second after Paul starts....he closes the gap between himself and Peter by 7 - 4 = 3m/s

So.....6 seconds after Peter starts, Peter has traveled 6 * 4 = 24 m

So....the number of seconds that it will take Paul to catch Peter = 24 /3 = 8 sec