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npetrous · May 12, 2009, 06:51 PM

How do I find the average speed when given two speeds in m/s but no distance?

Perito · May 12, 2009, 08:15 PM

It's not possible to calculate average speed with only this information. You're quite correct to question it.

jcaron2 · May 12, 2009, 09:27 PM

You need to know either the distances traveled at each speed (which you already said you DON'T know) or the amount of time for which each speed was maintained. If you know the time, the average speed is (s1*t1 + s2*t2)/(t1 + t2), where s1 and s2 are the speeds and t1 and t2 are the respective times. The times can be in any units (minutes, seconds, etc. or even relative units like "1/3 of the trip traveled at 10 m/s and 2/3 of the trip traveled at 20 m/s". In this last case the values of t1 and t2 would be 1/3 and 2/3).

aroraak · May 12, 2009, 09:45 PM

It is not possible to know average speed if you don't know distance traveled or time taken for each speed one parameter has to be known i.e either distance or time taken

rishabbest2009 · May 17, 2009, 09:28 AM

A train travels some distance with a speed of 30 km/h and returns with a speed of 45 km/h.
Calculate the average speed of the train?

Unknown008 · May 17, 2009, 10:06 AM

Let's say the distance is x. Total distance will be 2x.

Time for first trip = x/30
Time for second trip = x/45

Average speed = 2x/(x/30 + x/45) = 2x/(x/18) = 36 km/h

Next time, post your answer in another thread.

Zeejay · Sep 27, 2010, 10:12 AM

Originally Posted by Unknown008

Let's say the distance is x. Total distance will be 2x.

Time for first trip = x/30
Time for second trip = x/45

Average speed = 2x/(x/30 + x/45) = 2x/(x/18) = 36 km/h

Next time, post your answer in another thread.

I'm curious as to where you got your "18" in your solution.
The formula you should be using, because you are not given distance and time, is:
2*s1*s2/(s1+s2)

Therefore,
2*30*45/(30+45)= x
2,700/(75)= x
x= 36 km/h
Somehow you got the same answer, but your formula is random and doesn't fit the problem...

Unknown008 · Sep 27, 2010, 10:16 AM

It's not random... I simplified it on a line.

Total distance = 2x

Time for first trip = x/30

Time for second trip = x/45

Total time = x/30 + x/45 = x/18

Hence, average speed = $\frac{Total\ Distance}{Total\ Time} = \frac{2x}{\frac{x}{18}} = 36$ km/h

I work from the basis, that is:

$Average\ Speed = \frac{Total\ Distance}{Total\ Time}$

I don't have to then memorize some additional formulae.

And if you didn't see earlier, I assigned a variable x to be the distance. The formula you used is only a simplified version of the process which I did.

Let's derive your formula.

A car travels along a distance x with speed s1.
The same car comes back with a speed s2.

The total distance is = 2x
The total time = $\frac{x}{s_1} + \frac{x}{s_2} = \frac{xs_1 + xs_2}{s_1s_2} = \frac{x(s_1+s_2)}{s_1s_2}$

Average speed = $\frac{2x}{$ \frac{x(s_1 + s_2)}{s_1s_2}$} = \frac{2}{$ \frac{s_1 + s_2}{s_1s_2}$} = \frac{2s_1s_2}{s_1 + s_2}$

The method is the same, except that you used a formula derived from my method :)

Zeejay · Sep 27, 2010, 08:18 PM

Originally Posted by Unknown008

It's not random... I simplified it on a line.

Total distance = 2x

Time for first trip = x/30

Time for second trip = x/45

Total time = x/30 + x/45 = x/18

Hence, average speed = $\frac{Total\ Distance}{Total\ Time} = \frac{2x}{\frac{x}{18}} = 36$ km/h

I work from the basis, that is:

$Average\ Speed = \frac{Total\ Distance}{Total\ Time}$

I don't have to then memorize some additional formulae.

And if you didn't see earlier, I assigned a variable x to be the distance. The formula you used is only a simplified version of the process which I did.

Let's derive your formula.

A car travels along a distance x with speed s1.
The same car comes back with a speed s2.

The total distance is = 2x
The total time = $\frac{x}{s_1} + \frac{x}{s_2} = \frac{xs_1 + xs_2}{s_1s_2} = \frac{x(s_1+s_2)}{s_1s_2}$

Average speed = $\frac{2x}{$ \frac{x(s_1 + s_2)}{s_1s_2}$} = \frac{2}{$ \frac{s_1 + s_2}{s_1s_2}$} = \frac{2s_1s_2}{s_1 + s_2}$

The method is the same, except that you used a formula derived from my method :)

It's just that, the question the kid gave above, doesn't state Time, or Distance... Just speeds. 30km/h and 45km/h.
I'm not against your formula in any way at all, in fact I praise the precision behind it.
I'm curious, what's your IQ?

Unknown008 · Sep 27, 2010, 11:14 PM

I don't think that IQ is reliable in any way. I think that everyone has intelligence, but it is expressed in different forms. There are the artistic, social, maths, logical, etc.

And I prefer working as far as possible, from the basics. This will help you understand from where formulae were derived, how to even derive formulae that you may find useful.

Ash Paul · Jul 2, 2014, 05:33 AM

THX Unknown008... your answer was helpful to me!! :)