Question 1142180: What was their average speed?
Two youngsters were running as hard as they could. They averaged 6 miles per hour, and then had to rest. On the way back, they averaged only 4 miles per hour for the same distance. Not counting resting time, what was their average speed?
Found 2 solutions by Alan3354, Theo:
Answer by Alan3354(69443)   (Show Source):
You can put this solution on YOUR website! Two youngsters were running as hard as they could. They averaged 6 miles per hour, and then had to rest. On the way back, they averaged only 4 miles per hour for the same distance. Not counting resting time, what was their average speed?
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Avg speed of a round trip is
2*R1*R2/(R1 + R2)
= 2*6*4/(6+4)
= 4.8 mi/hr
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Looks like Theo is going for his PHD on this one.
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"rate of speed" is redundant, at best.
It's either rate, or speed.
Answer by Theo(13342)  (Show Source):
You can put this solution on YOUR website! r * t = d
r = rate
t = time
d = distance.
going,the rate if 6 miles per hour.
coming back, the rate is 4 miles per hour.
let t1 equal the time going.
let t2 equal the time coming back.
let t = t1 + t2 = total time going and coming.
let d = distance going and let d = distance coming back.
total distance going and coming back is therefore 2d.
going, the formula becomes 6 * t1 = d
coming back, the formula becomes 4 * t2 = d
in the first equation, solve for t1 to get t1 = d/6
in the second equation, solve for t2 to get t2 = d/4
since t = t1 + t2, then t = d/6 + d/4.
solve for t in terms of d to get t = 10d/24
the overall time going and coming is 10d/24
the overall distance going and coming is 2d.
rate * time formula becomes r * 10d/24 = 2d
solve for r to get r = 2d * 24/10d = 48d / 10d = 48/10 = 4.8
the overall average speed is 4.8 miles per hour.
this formula should hold regardless of the distance.
for example, assume the distance is 24 miles going and 24 miles coming back.
the formula for going and coming becomes 4.8 * t = 48.
solve for t to get t = 10 hours.
you have 10 = t1 + t2.
since t1 = d/6, then t1 = 24 / 6 = 4
since t2 = d/4, then t2 = 24 / 4 = 6
t1 + t2 = 4 + 6 = 10.
all the numbers are confirmed to be good.
your solution is that the overall average speed is equal to 4.8 miles per hour.
you don't have one fixed solution, such as a specific distance, or specific times, because you are dealing with 2 equations in 3 unknowns.
the 2 equations are 6 * t1 = d and 4 * t2 = d
the unknowns are t1, t2, and d.
your overall going and coming back equation is 4.8 * t = d
now you have 1 equation in 2 unknowns.
no single solution is possible.
you can fix the distance and calculate the time, or you can fix the time and calculate the distance, so your solution is dependent on specifying one of the unknowns so you can solve the other.
i did that above by specifying the distance to be 48 miles.
if i used anothe distance, i would have gotten another overall time and then proceeded to get the time required for going and the time coming back.
for example, assume the distance was 2.4 miles going and 2.4 miles coming back.
6 * t1 = 2.4 would get a t1 of 2.4 / 6 = .4
4 * t2 = 2.4 would get a t2 of 2.4 / 4 = .6
total time of t = t1 + t2 = .4 + .6 = 1 hour.
overall formula becomes r * 1 = 4.8
solve for r to get r = 4.8
different distance, different time, same overall rate of speed.
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