1. ## Average speed

Am I right in saying this is impossible to calculate? You need the average speed of the wagon rider...

A wagon train that is one mile long advances one mile at a constant rate. During the same time period, the wagon master rides his horse at a constant rate from the front of the wagon train to the rear, and then back to the front. How far did that wagon master ride?

*He does not simply go down and back. The wagon train is constantly moving forward

2. Il depends from where you are taking measurements : if you are using the wagon, then the wagon master rode 2 miles (1 to the end, 1 back).

If you use the Earth, then I think it should be necessary to know the wagon master's speed, but I'm tired right now ... did you try making a little drawing ?

3. Originally Posted by Bacterius
Il depends from where you are taking measurements : if you are using the wagon, then the wagon master rode 2 miles (1 to the end, 1 back).

If you use the Earth, then I think it should be necessary to know the wagon master's speed, but I'm tired right now ... did you try making a little drawing ?
Yes, but if it doesn't tell you the constant rate then you can assume it's anything, leading to an unanswerable question, or infinite one...yes? *confused*

4. Originally Posted by Mukilab
Yes, but if it doesn't tell you the constant rate then you can assume it's anything, leading to an unanswerable question, or infinite one...yes? *confused*
$\displaystyle V = \frac{D}{T}$. You only know $\displaystyle D$. You need at least $\displaystyle V$ or $\displaystyle T$ to answer your question (I think).
Where did you find this question ?

5. Originally Posted by Bacterius
$\displaystyle V = \frac{D}{T}$. You only know $\displaystyle D$. You need at least $\displaystyle V$ or $\displaystyle T$ to answer your question (I think).
Where did you find this question ?
Someone was trying to be snippy and see if I was 'smart enough' to help my friend with his maths homework.

But D is constant and recurring, so you do not really know D

6. Then you do not know anything, and you have three unknowns. Thus, this question cannot be answered precisely without more information.

7. Ok, thanks

8. Originally Posted by Mukilab
Am I right in saying this is impossible to calculate? You need the average speed of the wagon rider...

A wagon train that is one mile long advances one mile at a constant rate. During the same time period, the wagon master rides his horse at a constant rate from the front of the wagon train to the rear, and then back to the front. How far did that wagon master ride?

*He does not simply go down and back. The wagon train is constantly moving forward
Let the wagon train's speed be v1 and the wagonmaster's speed be v2, both in m/h and both relative to the earth. While the wagonmaster rides to the rear, his speed, relative to the wagon train, is v1+ v2 and he will take 1/(v1+v2) hours to do that. The wagon train will have moved forward a distance v1/(v1+v2) miles. So he will have ridden 1- v1/(v1+v2) miles. Riding back to the front, his speed relative to the wagon train is v2- v1 m/h and he will take 1/(v2- v1) hours. The wagon train will have moved forward v1/(v2-v1) miles so he will have ridden 1+ v1/(v2-v1) miles.

He will have ridden a total of 1- v1/(v1+v2)+ 1+ v1/(v2-v1)= 2- (v1/(v1+v2)+ v1/(v2- v1)= 2+ (-v1(v2-v1)+ v1(v1+v2))/((v1+v2)(v2-v1))= 2+ 2v1^2/(v2^2- v1^2).

It is impossible to give a precise number without knowing at least the wagonmaster's speed as a multiple of the wagon train's speed.

9. Originally Posted by Mukilab
Am I right in saying this is impossible to calculate? You need the average speed of the wagon rider...

A wagon train that is one mile long advances one mile at a constant rate. During the same time period, the wagon master rides his horse at a constant rate from the front of the wagon train to the rear, and then back to the front. How far did that wagon master ride?

*He does not simply go down and back. The wagon train is constantly moving forward
Suppose the wagon train is moving at approximately 1 inch per century. (It is moving.)
Suppose the wagon master can ride at a blazing speed of 1 mile per hour.
In 1 hour the wagon master will be at the rear of the wagon train, then (instantly changing direction) the wagon master will be at the front of the wagon train in 1 hour.
How far did the wagon master ride? Two miles (rounded to the nearest inch)
--
Suppose the wagon train is moving at EXACTLY 1 mile per hour.
Suppose the wagon master rides at 1 mile per hour.
In 1/2 hour the wagon master will be at the rear of the wagon train, then (instantly changing direction) the wagon master will be at the rear of the wagon train travelling at the same speed as the wagon train. The wagon master will NEVER get back to the front of the wagon train.

You need to know the speed of the rider relative to the speed of the wagon train; then you can determine the total distance the wagon master would ride.

Am I right in saying this is impossible to calculate?
YES

.

10. That's what I got the first time, because I skim read it I thought the wagon rider went at 1mph which would be impossible to catch up, until I re-read the question and realized it was even worse! :P