I don’t know about anyone else, but whenever I encounter a problem like this, involving rates, I immediately do a dimensional analysis. I just took a stab at the problem and got a contradiction. I think this is equivalent to what you’re saying.
And I think I’m dawning on the same realization you’re having, though I don’t exactly follow your analytical explanation in terms of “mappings”, and that is that I shouldn’t have had to start writing down equations to understand that the problem is impossible. So that’s why I want an intuitive understanding for why the problem is impossible. I don’t quite have that yet.
St, Tt, Dt: The average speed, time interval and distance for the total trip.
S1, T1, D1: The average speed, time interval and distance for the first half of the trip.
S2, T2, D2: The average speed, time interval and distance for the second half of the trip.
Now, running through the equations with the numbers plugged in gives me the strange equation: T1 + T2 = T1. This is a contradiction, because it is premised that T2 > 0 (which can be inferred from denying infinite speeds).
But I want an intuitive understanding, so I decided to start over using generic constants rather than specifying 40mph and 20mph in the problem. So I start with computing the total average speed for the trip (this comes from the definition of average speed):
St = (S1 T1 + S2 T2) / (T1 + T2)
We know that D1 = D2 (because they are both half the distance of the whole trip), and thus S1 T1 = S2 T2 (from the formulas for speed, time and distance), so therefore:
St = 2 S1 T1 / (T1 +T2)
T1 + T2 = (2 S1 / St) T1, if T1 + T2 != 0
All of these values need to be positive numbers, so S1 / St > 1⁄2. Otherwise, the problem yields a contradiction. The numeral 2 in the above equation comes from the fact that D1 is half the distance of Dt, so it seems likely that if the problem were stated differently such that D1 was a third of Dt, then it would need to be true that S1 / St > 1⁄3; and S1 / St > 1⁄4 if the distance is a fourth; and so on.
I don’t know about anyone else, but whenever I encounter a problem like this, involving rates, I immediately do a dimensional analysis. I just took a stab at the problem and got a contradiction. I think this is equivalent to what you’re saying.
And I think I’m dawning on the same realization you’re having, though I don’t exactly follow your analytical explanation in terms of “mappings”, and that is that I shouldn’t have had to start writing down equations to understand that the problem is impossible. So that’s why I want an intuitive understanding for why the problem is impossible. I don’t quite have that yet.
Okay, here’s my analysis:
The characters:
St, Tt, Dt: The average speed, time interval and distance for the total trip. S1, T1, D1: The average speed, time interval and distance for the first half of the trip. S2, T2, D2: The average speed, time interval and distance for the second half of the trip.
Now, running through the equations with the numbers plugged in gives me the strange equation: T1 + T2 = T1. This is a contradiction, because it is premised that T2 > 0 (which can be inferred from denying infinite speeds).
But I want an intuitive understanding, so I decided to start over using generic constants rather than specifying 40mph and 20mph in the problem. So I start with computing the total average speed for the trip (this comes from the definition of average speed):
St = (S1 T1 + S2 T2) / (T1 + T2)
We know that D1 = D2 (because they are both half the distance of the whole trip), and thus S1 T1 = S2 T2 (from the formulas for speed, time and distance), so therefore:
St = 2 S1 T1 / (T1 +T2)
T1 + T2 = (2 S1 / St) T1, if T1 + T2 != 0
All of these values need to be positive numbers, so S1 / St > 1⁄2. Otherwise, the problem yields a contradiction. The numeral 2 in the above equation comes from the fact that D1 is half the distance of Dt, so it seems likely that if the problem were stated differently such that D1 was a third of Dt, then it would need to be true that S1 / St > 1⁄3; and S1 / St > 1⁄4 if the distance is a fourth; and so on.