What is the average speed of a cyclist on a round trip journey?

[blaziken's_charizard]

1. The problem: A cyclist travels to work at an average speed of 3 m/s and returns home at an average of 9 m/s. Calculate the average speed for the whole journey.


2. Hints given by the teacher:
The answer is not 6 m/s & the answer is around [approximately] 4.5 m/s.

3. The distance was not given, if it is needed...

*m/s is meters per second.

 

[pmp!]

And I guarantee you that your teacher is right! It would be more fun if you think again for yourself. Focus on the definition of average velocity.

 

[berkeman]

*m/s is miles per second.

No, m/s is meters per second. Think about it -- if it were miles per second, that would be one FAST biker!

And on your problem, approach it with the definition of average speed. It's distance per time, right? So write the equation for the speed in terms of the distance and time. The distance is the same both coming and going, right? So what does that say about each time? And then how do you combine them to get the overall average?

 

[blaziken's_charizard]

Oh, yeah sorry about that. I made a mistake about putting miles instead of meters.

I still don't get what you guys are saying... :(

 

[berkeman]

I still don't get what you guys are saying... :(

Well, we're not going to do your homework for you -- that's against the PF rules. We can provide hints and pointers, though. Here's one pointer:

http://en.wikipedia.org/wiki/Speed

Now, try using some specific numbers and experimenting with the speed equation. What is the equation for average speed in terms of the distance and total time it takes to cover that distance? Just use the simplest equation from your text or the wikipedia article. Now make up some numbers, like say the trip to work is 1km (1000 meters) long. How long does it take for the biker to get to work? How long to get back home? Now what is the total distance for the round trip? And how long did a full round trip take? And what does that mean the overall average speed is for the round trip?

 

[dontdisturbmycircles]

edit:
---
nevermind I think that you can figure it out with the info berkeman gave you.

 

[blaziken's_charizard]

No, m/s is meters per second. Think about it -- if it were miles per second, that would be one FAST biker!

And on your problem, approach it with the definition of average speed. It's distance per time, right? So write the equation for the speed in terms of the distance and time. The distance is the same both coming and going, right? So what does that say about each time? And then how do you combine them to get the overall average?

It says that one time is faster than the other. In that, the first journey [to work] he was slower and the second journey [back home] he was faster...

...so if I were to assume that the distance from his house to his workplace [and vice-versa] is 18m and;

1) on his trip to work he took 6 sec.
2) on his trip back home, he took 2 sec.


...and if I also assume that the distance is 27m [for solid evidence];

1) on his trip to work he took 9 sec.
2) on his trip back home, he took 3 sec.

...it can be concluded that the time taken to go home is thrice the time he took to go to work.

So, therefore:

Total speed when heading to work is: 3 m/s
Total speed when heading back to home covering same distance: 9 m/s
Therefore, the final speed is thrice that of initial speed: 9/3 = 3
Average speed is: 3 m/s + 9 m/s = 12 m/s / 3 =

4 m/s

So yeah, 4 m/s is close enough to 4.5 m/s. I hope I'm right. Am I?

 

[cristo]

No, you're not! Thinking of it this way may help. The ratio of the outward speed to the inward speed is 1:3.

In a sense, you were right in thinking to treble the outward speed, but you forget to add on the bit that comes from the inward speed. And then, since you are taking an average.. you need to divide by something

[hint: if the speeds were equal, you would have (1*outward+1*inward/2)]

I don't know whether this helps, but its another way to look at it, if you can't see the answer yet!

 

[blaziken's_charizard]

Now that is even more confusing.

 

[Dorothy Weglend]

...so if I were to assume that the distance from his house to his workplace [and vice-versa] is 18m and;

1) on his trip to work he took 6 sec.
2) on his trip back home, he took 2 sec.

Good. Right. t = D/v

...and if I also assume that the distance is 27m [for solid evidence];

1) on his trip to work he took 9 sec.
2) on his trip back home, he took 3 sec.

Also good. Right again, t = D/v

...it can be concluded that the time taken to go home is thrice the time he took to go to work.

So, therefore:

Total speed when heading to work is: 3 m/s
Total speed when heading back to home covering same distance: 9 m/s
Therefore, the final speed is thrice that of initial speed: 9/3 = 3
Average speed is: 3 m/s + 9 m/s = 12 m/s / 3 =

4 m/s

So yeah, 4 m/s is close enough to 4.5 m/s. I hope I'm right. Am I?

Why did you ignore the distance in your final calculation?

Dorothy

 

[Kurdt]

All you need to assume is that the distance is some fixed constant. Secondly you find the equation for average speed from your notes or the recommended text or the internet and you put in the information you have. That is the distancewhich will be 2 times your constant and the time which is the sum of the initial speed over the distance and the final speed over the distance. The rest is simple algebra. Combined with the other posts this should be all the information you need as I've practically given you a verbal recipe.

 

[blaziken's_charizard]

Thanks guys, I think I *finally* solved it.