Which body has the higher average speed?

[shinokk]

Homework Statement

Two bodies travel the same distance. One body travels half of the distance with the speed V and the other half with the speed 2V. The second body travels half of the time with the speed V and the other half of the time with the speed 2V. Which of the bodies has the higher average speed?

Homework Equations

V= s/t
where V is speed, s is distance and t is time.

The Attempt at a Solution

The average speed of the first body is:
Vav1=((s/2)/t + (s/2)/(t/2))/2 = (s/2t + 2s/2t)/2= (3s/2t)/2 = (3/2*V)/2= 3/4*V

The average speed of the second body is:
Vav2= ((1/2*t*V + 1/2*t*2V)/t)/2 = (t*(1/2*V + V)/t)/2 = (1/2*V+V)/2= (3/2*V)/2= 3/4*V

Conclusion: The bodies have the same average speed.

Second try:

The average speed of the first body is:
Vav1 = (s/2)/t + (s/2)/(t/2) = s/2t + 2s/2t = (3s/2t)/2 = 3/2*V

The average speed of the second body is:
Vav2 = (1/2*t*V + 1/2*t*2V)/t = t*(1/2*V + V)/t = 1/2*V+V = (3/2*V)/2 = 3/2*V

But then this would only be: Vav= (V+2V)/2=3/2*V

 

[Doc Al]

No.

The average speed is distance over time. So figure out the total time each takes. (Don't try to directly average the speeds.)

 

[TSny]

Since each body never travels slower than V and spends some time traveling faster than V, the avg speed for either body should be greater than V.

 

[shinokk]

No.

The average speed is distance over time. So figure out the total time each takes. (Don't try to directly average the speeds.)

How? The values are not given in the problem statement.

Since each body never travels slower than V and spends some time traveling faster than V, the avg speed for either body should be greater than V.

Ahh, yes. Then what am I doing wrong? Perhaps the last division by 2 is unnecessary and the result would be 3/2*V for both?

 

[Doc Al]

How? The values are not given in the problem statement.

Use distance = speed*time to figure out the time in terms of what's given.

 

[Chestermiller]

You should follow Doc Al's advice. He knows what he is talking about.

Body 1:

Let s be the total distance. In terms of V and s, how much time does it take the body to cover the first half of the distance? In terms of V and s, how much time does it take the body to cover the second half of the distance? What is the total time. What is the average velocity?

Body 2:

Let T be the total time. In terms of V and T, how much distance does the body cover in the first half of the time? In terms of V and T, how much distance does the body cover in the second half of the time? What is the total distance covered? What is the average velocity?

 

[shinokk]

Use distance = speed*time to figure out the time in terms of what's given.

The solution in my book is:

Vav1= s/s/2v + s/s/4v = 4/3V
Vav2= (V*t/2 + 2V*t/2)/t = 3/2V

What I don't understand is how s/s/2v + s/s/4v equals 4/3V... I checked with Wolframalpha and it's correct.

 

[shinokk]

You should follow Doc Al's advice. He knows what he is talking about.

Body 1:

Let s be the total distance. In terms of V and s, how much time does it take the body to cover the first half of the distance? In terms of V and s, how much time does it take the body to cover the second half of the distance? What is the total time. What is the average velocity?

Body 2:

Let T be the total time. In terms of V and T, how much distance does the body cover in the first half of the time? In terms of V and T, how much distance does the body cover in the second half of the time? What is the total distance covered? What is the average velocity?

So, I followed your advices and I got this for body 1:
t1= s/2/v
t2=s/2/2v
From that I got that:
Vav1= s/(s/2/v + s/2/2v) and then this somehow equals 4/3V. I don't understand how, but it does according to WolframAlpha.

For body 2 I got:
s1= v*t/2
s2= 2v*t/2
stotal = 3/2v*t
Vav2= (3/2v*t)/t= 3/2v

Can you explain how Vav1= 4/3V? I tried calculating it myself but I can't get the correct solution.

 

[haruspex]

So, I followed your advices and I got this for body 1:
t1= s/2/v
t2=s/2/2v
From that I got that:
Vav1= s/(s/2/v + s/2/2v) and then this somehow equals 4/3V. I don't understand how, but it does according to WolframAlpha.
Can you explain how Vav1= 4/3V? I tried calculating it myself but I can't get the correct solution.

Please post your working.

 

[TSny]

So, I followed your advices and I got this for body 1:
t1= s/2/v
t2=s/2/2v

Write these as

$t_1 = \frac{s/2}{v} = \frac{s}{2}\frac{1}{v} = \frac{s}{2v}$

$t_2 = \frac{s/2}{2v} = \frac{s}{2}\frac{1}{2v} = \frac{s}{4v}$

So, what do you get for the total time if you add these two fractions and combine into one fraction?

 

[shinokk]

Please post your working.

t = s/V
Since the body travels half the distance with the speed V then:
t1= (s/2)/V
For the second half of the distance the speed is 2v then:
t2= (s/2)/2V

But now I see that it would be easier to write t1=s/V and t2= s/2v and later, when calculating the average speed, to just use 2s.

Write these as

$t_1 = \frac{s/2}{v} = \frac{s}{2}\frac{1}{v} = \frac{s}{2v}$

$t_2 = \frac{s/2}{2v} = \frac{s}{2}\frac{1}{2v} = \frac{s}{4v}$

So, what do you get for the total time if you add these two fractions and combine into one fraction?

3s/4v :(

So, maybe this:
t1= S/V
t2= S/2V
Vav1= 2S/(S/V + S/2V)

And I don't know how to calculate this without WolframAlpha, but it gives the correct solution (4/3V). Could any of you shed some light on how it's calculated?
So, in the end it's the second body that has the higher average speed. Vav1= 4/3V and Vav2=3/2V

 

[haruspex]

Vav1= 2S/(S/V + S/2V)
And I don't know how to calculate this without WolframAlpha

Oh dear. Is WolframAlpha going to produce a generation of scientists that can't manipulate algebra? But then, I still remember slide rules and log tables!
2S/(S/V + S/2V)
Multiply top and bottom by 2V:
4SV/(2S + S)
Divide top and bottom by S:
4V/(2 + 1) = 4V/3

 

[shinokk]

2S/(S/V + S/2V)
Multiply top and bottom by 2V:
4SV/(2S + S)
Divide top and bottom by S:
4V/(2 + 1) = 4V/3

Ohh, snap! It's so obvious that I can't believe I didn't see that. o.o