# Average Speed?

1. Sep 23, 2004

### hytuoc

You drive on Interstate 10 from San Antonio to Houston, half the distance at 55km/h and the other half at 90km/h. Waht is your average speed from San Antonio to Houson?
How do I do this?

2. Sep 23, 2004

### DrWarezz

To find the average of something, you simply add up all the numbers you have, then divide by how many you have, as an answer to your question:

(55+90)/2 = 72.5

:)
[r.D]

3. Sep 23, 2004

### Tide

The average speed is the total distance travelled divided by the total time to make the trip.

4. Sep 23, 2004

### DrWarezz

Tide, that's the overall speed. Which would equal the average speed. So, you're not necessarily wrong, however, niether the time nor the distance are specified in the question. So, my answer is the one to go for :)

[r.D]

5. Sep 23, 2004

### robphy

That's the problem with the lack of a good explanation of "average velocity" in physics textbooks.

"average velocity" is a time-weighted-average of velocities.
$$v_{av}=\frac{v_1 \Delta t_1 + v_2 \Delta t_2}{\Delta t_1+\Delta t_2}$$
Only then does it logically follow that
$$v_{av}=\frac{\Delta x}{\Delta t}$$

Generally speaking (i.e., for non-constant acceleration), the straight-average velocity has no physically interesting application.

edit:
"average speed" is a time-weighted-average of speeds.
$$s_{av}=\frac{|v_1 \Delta t_1| + |v_2 \Delta t_2|}{\Delta t_1+\Delta t_2}$$
Only then does it logically follow that
$$s_{av}=\frac{d_{total}}{\Delta t}=\frac{d_{total}}{t_{total}}$$
where $d_{total}$ is the total-distance travelled.

6. Sep 23, 2004

### DrWarezz

Thanks robphy.
However, my solution answers the authors question, correct?

:)

[r.D]

7. Sep 23, 2004

### robphy

Since v1t1=v2t2, I get 68.276 km/hr.

8. Sep 23, 2004

### DrWarezz

\ Hmm..
Well, I'm guessing by the simplicity of the question, that such a method is not necessary. Instead, the teacher that set this homework :P is just wanting to see the use of averages. :) hehe.

So, to the author: If this is just some homework, and you haven't done averages at school yet (I'm assuming you're at school), then go for my answer. But, if you're, eg, at college or something, definitely go for robphys' solution :D

best of luck - and thanks for the insight robphy, I am only a young school student (15 yrs old), so, your solution is quite new to me. But, I've bookmarked it, and will surely use it in the future :D Thanks.

[r.D]

9. Sep 23, 2004

### ZapperZ

Staff Emeritus
You cannot apply the statistical definition of "average" to average speed. It is DEFINED, as stated, as the total distance travelled divided by the total time taken. Even though the total distance and time taken here are not given, this problem is still solvable if one stick to solving it symbolically in the beginning and not plug in numbers right away.

Zz.

P.S. This thing is posted in the wrong section of PF. We do have a Homework Help section.

10. Sep 23, 2004

### Tide

Well, you could certainly do that but consider this. Suppose you traveled at 50 km/h for the entire trip except that over the last meter you gunned it and traveled at the speed of light. Would your average speed have been half the speed of light?

11. Sep 23, 2004

### krab

The reciprocal of the average speed is the average of the reciprocals:
$$\overline{v}={d\over t}={d\over t_1+t_2}={d\over{d\over 2v_1}+{d\over 2v_2}}={2v_1v_2\over v_1+v_2}$$

12. Sep 23, 2004

### Tide

More generally, if the fraction of the total distance traveled at speed $v_1$ is $f$ with the remainder at speed $v_2$ then the average time is
$$\frac {v_1 v_2}{f v_2 + (1-f)v_1}$$

13. Sep 23, 2004

### ArmoSkater87

The solution offered in the very biginning,
$$\overline {v} = {v_0 + v_f \over 2}$$ only works if assuming constant acceleration.

14. Sep 23, 2004

### ArmoSkater87

This problem in general is very dumb because it implies an instantaneous change in speed from 55km/h to 90km/h.