A question about average speed

[murshid_islam]

If we travel s₁ distance in t₁ time and s₂ distance in t₂ time, is the average speed \frac{s_1 + s_2}{t_1 + t_2} or \frac{\frac{s_1}{t_1} + \frac{s_2}{t_2}}{2}

 

[Doc Al]

The first one: (total distance)/(total time).

 

[murshid_islam]

The first one: (total distance)/(total time).

Is there any reason why it isn't the second one or is it just defined that way?

 

[Doc Al]

It's just what is meant by average speed.

 

[Mark44]

To follow up on what Doc Al said, average speed is defined as the total distance covered divided by the amount of time elapsed.

Here's an old problem that shows why the "average of averages" (as in your second formula) doesn't work.

On a certain road there is a hill that the road goes up and then down the other side. The uphill side is one mile and the downhill side is one mile. If I drive up the hill at an average speed of 30 mph, how fast do I need to drive down the other side to average 60 mph for the two miles?

 

[HallsofIvy]

The "average of two numbers" is NOT, in general, the same as "average speed".

 

[Mentallic]

I can see why you thought it might be the second one,

since \frac{s_1}{t_1}=v_1 and \frac{s_2}{t_2}=v_2 and the average of two speeds is \frac{v_1+v_2}{2} then in conclusion the average speed over the whole distance would be your second one intuitively.

While it seems plausible, it's not correct.

There is a difference in these statements, for your problem, we want the average speed over the whole distance while for this problem, we just want the average of the two velocities. The two aren't the same because if we take s_2=1 but t_2 really small (approaching zero) such that v_2 is really large (approaching infinite), and s_1=t_1=v_1=1 then the average speed by the method above would be (1+\infty)/2=\infty but what it should really be is (1+1)/(1+0)=2 since even though it travels really fast on the second journey, that journey is completed in such a short time that it doesn't mean the entire journey was completed in an infinitely small time (since we took a finite time to complete the first leg of the journey).

 

[murshid_islam]

Here's an old problem that shows why the "average of averages" (as in your second formula) doesn't work.

On a certain road there is a hill that the road goes up and then down the other side. The uphill side is one mile and the downhill side is one mile. If I drive up the hill at an average speed of 30 mph, how fast do I need to drive down the other side to average 60 mph for the two miles?

I'm sorry, I still don't get it. Why does the second formula, i.e., \frac{\frac{s_1}{t_1} + \frac{s_2}{t_2}}{2} or \frac{v_1 + v_2}{2} not work here? If we use that formula, we get the velocity required downhill to be 90 mph. On the other hand, if I use the first formula, i.e., \frac{s_1 + s_2}{t_1 + t_2}, we have,

s₁ = s₂ = 1 mile
v₁ = 30 mph
t₁ = 1/30 h

Now, average speed = \frac{s_1 + s_2}{t_1 + t_2}
60 = \frac{1+1}{\frac{1}{30} + t_2}

t_2 = 0

Then, v_2 = \frac{1}{0}

 

[Doc Al]

I believe Mark44 made an error in his problem statement. To average 60 mph, you'd have to complete the entire 2 mile trip in 2 minutes. But if you traveled the first half at 30 mph you've already used up 2 minutes. You'd have to travel the second mile in 0 time! (Infinite speed!) Which is impossible, of course.

But the formula works just fine--it tells you that you need to cover that second mile in 0 time.

 

[murshid_islam]

I believe Mark44 made an error in his problem statement. To average 60 mph, you'd have to complete the entire 2 mile trip in 2 minutes. But if you traveled the first half at 30 mph you've already used up 2 minutes. You'd have to travel the second mile in 0 time! (Infinite speed!) Which is impossible, of course.

But the formula works just fine--it tells you that you need to cover that second mile in 0 time.

But why exactly doesn't the second formula, i.e., \frac{v_1 + v_2}{2} work here?

(I know that the books tell us that the formula for average speed is total distance divided by total time.I just want to know why it isn't the average of the two speeds, i.e., \frac{v_1 + v_2}{2}.)

 

[Doc Al]

But why exactly doesn't the second formula, i.e., \frac{v_1 + v_2}{2} work here?

Well, that formula gives the average of two speeds but it's not the average speed.

I'm sorry, I still don't get it. Why does the second formula, i.e., \frac{\frac{s_1}{t_1} + \frac{s_2}{t_2}}{2} or \frac{v_1 + v_2}{2} not work here? If we use that formula, we get the velocity required downhill to be 90 mph.

But 90 mph isn't the right answer. If the second mile is traveled at a speed of 90 mph, the time it takes would be 1/90 hour. That would make the total time equal to:
1/30 + 1/90 = 4/90 hour.

But that makes the average speed equal to:
(2 miles)/(4/90 hour) = 45 mph (not 60, like it was supposed to be).