<PRE><B>Question 81960</B>
let t=time (in hours) after truck #2 leaves until it catches truck #1

distance(d)=rate(r)times time(t)or d=rt; t=d/r and r=d/t

When truck #2 leaves at 11:00am, truck #1 has already travelled (50 mph)(2 hrs)=100 mi

distance truck #1 travels =100+50t
distance truck #2 travels=75t 

Now we know that when the distance truck #1 travels equals the distance truck #2 travels, then truck #2 will have caught up with truck #1. So:

100+50t=75t subtract 50t from both sides
100+50t-50t=75t-50t  collect like terms
25t=100
t=4 hrs  --------------time it takes truck #2 to catch truck #1

11:00am plus 4 hrs=3:00pm----------time when truck #2 catches up with truck #1

CK

truck #1 travels 100+4(50)=300mi
truck #2 travels 4(75)=300mi

Hope this helps----ptaylor

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