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?

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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

Tom D. | Very patient Math Expert who likes to teachVery patient Math Expert who likes to te...

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

Runners are running towards each other so relative speed of runners is 5 + 7 = 12 mph

Time to travel 8 mlies with 12 mph speed = 8/12 = 2/3 hrs = 40 miniutes.

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