klreed2 said:
Is it as simple as it looks?
Probably not. This is a standard question designed to make you think, rather than assume.
You travel distance D at a constant velocity v1. You return traveling distance D at a constant velocity v2. What is your average velocity for entire trip of distance 2D?
If v1 > v2, you will spend more time at v2, travelling the same distance.
We know this.
Distance = Rate * Time
We'll need the Average total rate, so Solve for Rate
Rate = Distance/Time
Reword it for what we need.
Average Rate = (Total Distance) / (Total Time)
We are given this, with the introduction of the two Time variables.
D = v1*t1
D = v2*t2
Solve these for t.
D/v1 = t1
D/v2 = t2
Total Distance = 2D
Total Time = t1 + t2 = D/v1 + D/v2
Average Rate = (2D)/(D/v1 + D/v2) = 2D/[(D)*(1/v1 + 1/v2)] = 2/[(v2 + v1)/v1*v2] = (2*v1*v2)/(v1+v2)
Not quite (v1 + v2)/2, is it?
Let's see if it makes sense:
D = 50 miles
v1 = 50 mph
v2 = 100 mph
t1 = 50 miles / 50 mph = 1 hour
t2 = 50 miles / 100 mph = ½ hour
Total Distance / Total Time = (2*50 mi)/(1 hr + ½ hr) = 200 mi / (3/2) hr = (200/3) mph = (66 + 2/3) mph.
Since we spent 2/3 of our time at 50 mph, it makes sense that the average rate would be on the low side of the simple average, 75 mph.
Does the formula work?
(2*v1*v2)/v1+v2 = (2*50*100)/(50 + 100) = 10000/150 = 1000/15 = 200/3 -- Check!