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# How long after they start will the runners meet?

Two runners, starting at the same time, run towards each other from opposite ends of an 8 mile trail. One runner is running at a rate of 5 mph, and the other is running at a rate of 7 mph. How long after they start will the two runners meet?

r_1 = 5 mph
r_2 = 7 mph

d_1 = distance first runner has run
d_2 = distance second runner has run

d_1 + d_2 = 8 miles

t_1 = time first runner has been running
t_2 = time second runner has been running

t_1 = t_2 = t = time they will meet

d_1 = r_1 * t = 5 t
d_2 = r_2 * t = 7 t

d_1 + d_2 = 8

5 t + 7 t = 8

t (5 + 7) = 8

t = 8/12 = 2/3 hours = 40 minutes
Hi again Donald;
(7 miles/hour)+(5 miles/hour)=12 miles/hour
distance=8 miles
distance=(rate)(time)
(distance)/(rate)=time
(8 miles)/(12 miles/hour)=time
Let's first note the unit of //hour is the same as the unit of hour in the numerator.
[(8 miles)/(12 miles)]hours=time
Let's note that the unit of miles in the numerator and denominator cancel...
[(8 miles)/(12 miles)]hours=time
(8/12)hours=time
(2/3)hours=time
If you want minutes...
[(2/3)hours][(60 minutes)/(1 hour)]=time
[(2/3)hours][(60 minutes)/(1 hour)]=time
(2/3)(60)minutes=time
(120/3)minutes=time
40 minutes=time
A good way to think about this is that both runners are contributing to the 8 mile meeting.  We don't care where they will meet in this problem (although you can figure that out too).  Therefore simply consider their combined rates (Rate=R5+R7) producing two distances (d5 & d7) which will ultimately total 8 miles.  Since Rate*Time=Distance:

Time=Distance/Rate = 8mi/12mph = 2/3hr = 40minutes

You could now plug this total time back in for each runner to see how far they ran

d5=Time*R5 = 2/3 hour*(5miles/hour)= 10/3 miles
d7=Time*R7 = 2/3 hour*(7miles/hour)= 14/3 miles

Note: d5+d7=8miles