G Guest Guest Nov 1, 2005 #1 If Sue leaves @ 9AM going 28 MPH, and John leaves 2 hours later @ 48MPH at what time will John be caught up with Sue?
If Sue leaves @ 9AM going 28 MPH, and John leaves 2 hours later @ 48MPH at what time will John be caught up with Sue?
G Gene Senior Member Joined Oct 8, 2003 Messages 1,904 Nov 1, 2005 #2 distance = velocity times time. T is the time of catching. Sue goes T-9 hours at 28 MPH John goes T-11 hours at 48 MPH 28*(T-9) = 48*(T-11)
distance = velocity times time. T is the time of catching. Sue goes T-9 hours at 28 MPH John goes T-11 hours at 48 MPH 28*(T-9) = 48*(T-11)
